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Detailed Design Study of North Java Corridor Flyover Project

BALARAJA FLYOVER DETAILED DESIGN

SUBSTRUCTURE

Katahira & Engineers International

8. DETAILED DESIGN OF COPING

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SUBSTRUCTURE


Notes on General Design and Location Features (1) Maintainability (AASHTO LRFD Section 2.5.2.3) Areas around bearing seats and under deck joints shall be designed to facilitate jacking, repair and replacement of bearings and joints.

Jacking points shall be indicated on the plans and the structure shall be designed for jacking forces.

The design forces for jacking in service shall not be less than 1.3 times the permanent load reaction at the bearing adjacent to the point of jacking (AASHTO Article 3.4.3). The bridge shall be assumed to be closed to traffic during jacking operations.

(2) Criteria for Deflection (AASHTO LRFD Section 2.5.2.6.2) The following deflection limits shall be considered:

• Vehicular load on cantilever arms………………………………………….Span/300

(3) Load Factors for Construction Loads (AASHTO LRFD Section 3.4.2) Load factors for the weight of the structure and appurtenances shall not be taken as less than 1.25.

The load factor for construction loads and for dynamic effects shall not be less than 1.5.

Notes on Structural Analysis and Application of Design Vehicular Loads (1) Equivalent width of Deck Overhang (AASHTO LRFD Table 4.6.2.1.3-1) The width of the equivalent strip of deck overhangs (cantilever slabs) shall be as follows:

Width of primary strip (mm) = 1140 + 0.833X …………………………(Equation 1)

where:

X = distance from load to point of support

Given that the deck overhang on the coping is not a continuous slab, Equation 1 above shall be modified to account for edge loading in accordance with AASHTO LRFD Article 4.6.2.1.4 as follows:

Width of primary strip (mm) = 570 + 0.416X …………………………..(Equation 1A)

(2) Application of Load (AASHTO LRFD Article 3.6.1.3)

The design truck shall be positioned transversely such that the center of any wheel is not closer than:

• For the design of deck overhang – 300mm from the face of the curb or railing

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Notes on Flexural Design (1) Loads and Load Combinations The loads and load combinations are taken from Section 6 and summarized in the following pages. Both ultimate limit state and serviceability limit state combinations were included in the design in accordance with the Design Criteria.

(2) Ultimate Moment Capacity (AASHTO LRFD Section 5.7) Ultimate moment capacity of reinforced concrete beams is determined in accordance with AASHTO LRFD Article 5.7.3.2.3 as follows.

The nominal flexural resistance of a singly reinforced beam without compression reinforcement is given as:

⎟⎠⎞

⎜⎝⎛ −⋅⋅=

2adfAM sySn ……………………………………………………(Equation 1)

where:

As = area of non prestressed tensile reinforcement

fy = yield strength of reinforcing bars

ds = distance from extreme compression fiber to the centroid of non prestressed tensile reinforcement

a = depth of equivalent stress block, 1β⋅c (Equation 2)

c = distance from extreme compression fiber to the neutral axis

1β = stress block factor

Assuming rectangular section behavior and yielding of reinforcement:

bffA

cc

yS

⋅⋅⋅

⋅=

185.0 β……………………………………………………………(Equation 3)

where:

b = width of rectangular section

fc = compressive strength of concrete

Substituting Equation 2 and 3 into Equation 1 gives:

⎟⎟⎠

⎞⎜⎜⎝

⎛⋅⋅

⋅⋅−⋅⋅=

bffA

dfAMc

ySsySn 85.02

1………………………………………(Equation 4)

Rearranging Equation 4 and dividing all terms by ( bd s ⋅2 ) gives:

0.085.02 222

22

=⋅

+⋅⋅

−⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅⋅⋅⋅

bdM

bdfA

bdffA

s

n

s

yS

sc

yS ………………………………..(Equation 5)

Defining terms:

bdA

s

S

⋅=ρ

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y

c

ff

M⋅

=85.0

φU

nM

M =

where:

MU = applied ultimate moment from factored loads

φ = strength reduction factor for flexure

Substituting defined terms into Equation 5 and dividing through by fy gives:

0.02 2

2

=⋅⋅⋅

+−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅ ys

U

fbdM

M φρρ

……………………………………..(Equation 6)

Defining terms:

ys

U

fbdMR

⋅⋅⋅= 2φ

Substituting defined terms into Equation 6 gives:

0.02

2

=+−⎟⎟⎠

⎞⎜⎜⎝

⎛⋅

RM

ρρ…………………………………………….…..(Equation 7)

Solving the quadratic Equation 7 gives:

⎟⎟⎠

⎞⎜⎜⎝

⎛ ⋅−−⋅=

MRM 211ρ ……………………………………………....(Equation 8)

Equation 8 gives directly the percentage of reinforcement required to resist the applied factored loads.

The Ultimate Moment Capacity of reinforced concrete columns is determined using the computer program PCACOL. This is based on ACI-95 and is consistent with the requirements of AASHTO LRFD.

(3) Service Moment Capacity (AASHTO LRFD Section 5.7) The solution for the analysis of reinforced concrete sections in flexure under no axial loading and no compression reinforcement is derived from:

• Linear stress-stress relations

• Plane sections remain plane under flexure

• Equilibrium of internal forces (compressive force in concrete is equal the tensile force in the reinforcement)

With reference to the illustration below for a rectangular section:

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Defining terms:

C = Compressive force in concrete

T = Tensile force in reinforcing steel

de = Effective depth from extreme compressive fiber to centroid of tensile reinforcement

c = Depth of concrete in compression

b = Width of rectangular section

cε = Compressive strain in concrete at extreme compressive fiber

sε = Tensile strain in reinforcement

cE = Modulus of elasticity of concrete

sE = Modulus of elasticity of reinforcement

α = Modular ratio, C

S

EE

=α

Establishing equilibrium of forces gives the following:

CT = ……………………………………………………………….………..(Equation 1)

SSS EAT ⋅⋅= ε …………………………………………………….………..(Equation 2)

cc EcbC ⋅⋅⋅

= ε2

…………………………………………………….………(Equation 3)

From considerations of compatibility of strains:

cdc e

sC

−=

εε………………………………………………………….……...(Equation 4)

C

S

EE

=α ……………………………………………………………….……..(Equation 5)

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Substituting Equation 2 to Equation 5 into Equation 1 gives:

αεε S

e

SSSS

Eccd

cbEA ⋅⋅−

⋅⋅

=⋅⋅2

………………………………….……….(Equation 6)

Dividing all terms of Equation 6 by ( )SS E⋅ε and rearranging gives:

0.02

2

=⋅−⋅+⋅⋅

eSS dAcAcbα

……………………………………….………..(Equation 7)

Solving the quadratic Equation 7 gives:

αα

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛−

⋅⋅⋅+

=b

AdAbAc

SeS

S22

………………………………………….(Equation 8)

Equation 8 gives directly the depth of concrete in compression, c, for a given area of reinforcing steel As.

The lever arm of the reinforcing steel, z, with respect to the centroid of the compressive force in the concrete is then obtained from:

3cdz e −= …………………………………………………………….………..(Equation 9)

The stress in the reinforcement, fs, can then be determined from:

zAMfS

Ss ⋅= …………………………………………………………….………..(Equation 10)

where:

MS = applied serviceability limit state moment from factored loads.

For beams with multiple layers of tensile reinforcement, As1, As2, As3, .......Asn, located at effective depths d1, d2, d3, ......dn, Equation 8 is modified as follows:

αα

⋅⎟⎟⎠

⎞⎜⎜⎝

⎛−

⋅⋅+

=b

ABbAc

TT22

…………………………………………….…..(Equation 8A)

where:

AT = Total area of tensile reinforcement, SnSSST AAAAA ............321 +++=

B = dnAdAdAdA SnSSS ⋅+⋅+⋅+⋅ ............321 321

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AASHTO Article 5.7.3.4 requires that components shall be so proportioned that the tensile stress in the mild steel reinforcement at the service limit state does not exceed fsa, determined as:

( )y

c

sa fAd

Zf ⋅≤⋅

= 6.031

where:

dc= depth of concrete from extreme tension fiber to center of bar (mm)

A = area of concrete having the same centroid as the principal tensile reinforcement and bounded by the surfaces of the cross-section and a straight line parallel to the neutral axis, divided by the number of bars (mm2)

Z = crack width parameter (N/mm)

For moderate exposure conditions, Z shall not exceed 30000 N/mm.

The Design Criteria established for the project (based on current Indonesian Standards and BMS) specifies that the allowable stress of reinforcing bars in tension shall be 0.5fy or 170MPa, whichever is smaller. (Design Criteria Table 2.4.2-2)

Given that for Grade 40 reinforcement fy = 390MPa, the 170MPa allowable stress implies a limit of 0.43fy.

The Design Criteria are therefore considered more onerous and will be applied for the serviceability checks on the coping.

(4) Limits for Reinforcement (AASHTO LRFD Article 5.7.3.3)

Maximum Reinforcement

The maximum area of longitudinal reinforcement for RC columns shall be such that:

42.0≤ed

c

where:

c = distance from extreme compression fiber to the neutral axis

bffA

cc

yS

⋅⋅⋅

⋅=

185.0 β

de = distance from extreme compression fiber to the centroid of non prestressed tensile reinforcement

1β = stress block factor

fy = yield strength of reinforcing bars

fc = compressive strength of concrete

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Minimum Reinforcement

The amount of tensile reinforcement shall be adequate to develop a factored flexural resistance, Mr, at least equal to the lesser of:

• 1.2 times the cracking moment, MCR , determined on the basis of elastic stress distribution and the modulus of rupture, fr, of the concrete:

cr ff ⋅= 63.0 in MPa

For monolithic construction:

rCCR fSM ⋅=

where:

Sc = section modulus for the extreme fiber of the section where tensile stress is caused

• 1.33 times the factored moment required by the applicable ultimate load combinations

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Notes on Shear Design – Reinforced Concrete Coping

(1) General Refer to Notes on Shear Design – Reinforced Concrete for the design approach.

(2) Column Connections (AASHTO LRFD Article 5.10.11.4.3)

The nominal shear resistance, Vn, provided by the concrete in the joint of a frame or bent in the direction under consideration, shall satisfy:

cn fdbV ⋅⋅⋅≤ 0.1

(3) Brackets and Corbels (AASHTO LRFD Article 5.13.2.4) The requirements of AASHTO are satisfied in the design of the components of the pier coping that can be considered as brackets and corbels.

(4) Beam Ledges (AASHTO LRFD Article 5.13.2.5) The requirements of AASHTO are satisfied in the design of the components of the pier coping that can be considered as beam ledges.

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Detailed Design Study of North java Corridor Flyover Project


SUBSTRUCTURE


8/8/2006

Notes on Joint Conection Design References :

R1. SEISMIC DESIGN AND RETROFIT OF BRIDGES – M.J.N Priestley, F. Sieble, G.M. Calvi. R2. SEISMIC DESIGN OF REINFORCED CONCRETE BRIDGES – Yan Xiao

Moment-Resisting Conection Between Column and Beam Connections are key elements that maintain the integrity of overall structure, they should be designed carefully to ensure the full transfer of seismic forces and moments. Because of their importance, complexity, and difficulty of repair if damaged, connections are typically provided with a higher degree of safety and conservativeness than column or beam members. Current AASHTO-LRFD Code do not provide specific design requirements for joint, except requiring the lateral reinforcement for columns to be extended into column/footing or column cap beam joint. A new design approach recently developed by Priestly & Calvi guidelines design is summarized bellow: Design Force In moment-resisting frame structures, the force transfer typically results in sudden changes (magnitude and direction) of moments at connections. Sudden moment change cause significant shear forces. Thus, joint shear design is the major concern of column and beam connection, as well as that longitudinal reinforcement of beam and column are to be properly anchored or continued through the joint to transmit. For seismic design, joint shear force can be calculated based on the equilibrium condition using force generated by maximum plastic moment acting on the boundary of connections. Momen Redistribution of Design Action

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8/8/2006

Beam Design for Flexure and Shear for beam are follow AASHTO LRFD Section 5.7 Flexure Design and AASHTO LRFD Section 5.8 Shear Design Truss mechanism contribution for beam is illustrated by twin-column bent of figure bellow :

At section 1-1

hence

………………………………………5.78a(R1)

At section 2-2

hence

………………………………………5.78b(R1)

P Fp Vcol.1−:=

Vp 0.85 Fp Vcol.1−( ) tan α( ):=

P' Fp Vcol.2−:=

Vp 0.85 Fp Vcol.2−( ) tan α( ):=

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8/8/2006

Design of Beam-Column Joint Shear Force in Beam-Column Joint The traditional approach for investigating force transfer in beam-column joint has been based on an assessment of the joint shear force developed from equilibrium considerations of the member force acting at the joint boundary. Consider knee joint in twin column P6 where column moment is overstrength, corresponding to plastic moment capacity in accordance with section 7 Column Design the column is considered as an independent member extending to the top of joint, with the influence of the beam or beams represented by force T, C, and VB applied to this independent member.

The overstrength moment Mo continues to increase above the level of the beam soffit until the line of action of the beam force C. The moment slope reverse under this force, decreasing to zero at the height of the upper stress resultant T. Note that an incremental moment decrease ΔM is shown at the beam lower stress resultant due to moment provided by the beam shear.

…………………………………………………………5.81a(R1)

…………………………………………………………..5.83(R1)

In similar fashion, the force acting on beam, considered to extend through the joint given

………………………………………………………..5.83(R1)

ΔM Vbhc2

⋅:=

VjhMohb

:=

VjvVjh hb⋅

hc:=

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8/8/2006

Knee Joints Knee joints are the most common type of joint occurring in multi column bridge bents when transverse response is considered. Equilibrium under closing and opening moment are represented in Figures bellow :

In these figures, the beam tensile, compressive and shear stress resultants are indicate by Tb, Cb, and Vb, with Tc, Cc, and Vcol being corresponding force for column. Axial forces Pc and Pb are present in the column and beam, respectively, and prestress force F is shown, which will, of course, be zero if the cap bem is reinforced conventionally. Moment Mb and Mc on the joint boundaries induce the flexural stress resultant note above. Equilibrium equations governing the relationships between the various force are summarized (R1) bellow :

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8/8/2006

Nominal Shear Stress The nominal shear stress in beam-column joints can be found directly from the joint shear force as :

………………………………………………………5.87(R1) Where bje is the effective width of the joint.

Design of Uncracked Joint Joint can be conservatively designed based on elastic theory for not permitting cracks. In this approach, the principal tensile stress within a connection is calculated and compared with allowable tensile strength . The principal tensile stress, pt and principal compression stress, pc with a simple Mohr’s circle analysis for stress shows that nominal principal stresses in joint region are given :

≤ 0.3 f’c

………………………………………5.89(R1,2)

where :

and ……………………………………………………………5.90(R1)

vjhVjh

bje hc⋅:=

vjvVjv

bje hb⋅:=

pcfv fh+

2

fv fh−

2

⎛⎜⎝

⎞⎟⎠

2

vj2

++:=

ptfv fh+

2

fv fh−

2

⎛⎜⎝

⎞⎟⎠

2

vj2

+−:= ft 0.29 f'c MPa⋅≤

fhPb

bb hb⋅:=

fvPc

bje hc 0.5 hb⋅+( )⋅:=

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8/8/2006

Mechanism of Force Tranfer in Cracked Joints When principal tension stresses excided the joint tension strength, cracking occurs and the force transfer from beam to column implied by equilibrium considerations can no longer be based on assumptions of isotropic material performance. Typical pattern of cracks developed in knee joint under closing and opening moments.

a. Closing Moment The amount of transverse hoop reinforcement required to provide anchorage of column reinforcement after splitting crack develop can be calculated by a shear friction approach.The hoop reinforcement should not excided that corresponding to a strain εσ=0.0015, since higher strains appear to result in excessive dilation of circumferential crack with reduced efficiency of the shear friction mechanism. A required volumetric ratio of transverse reinforcement to avoid anchorage failure is :

…………………………………………………….5.92(R1) Asc = Total area of column longitudinal reinforcement.

foyc = is overstrength stress in the column reinforcement, including strain hardening and yield

overstrength. = 1.4 fy fsh = 0.0015 Es again, the minimum requirement should be satisfied :

……………………………………………5.96(R1)

ρ s10.46 Asc⋅ foyc⋅

Dr la⋅ fsh⋅:=

ρ smin

0.29fc

MPa⋅ MPa⋅

fy:=

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8/8/2006

b. Opening Moment Three mechanisms to avoid the potential failure under opening moment, as show

The solution (a) and (b) is likely to cause unacceptable congestion and would require each of the tails of the column bars to be anchored with the resisting force of not less than 0.0033Ab fy . Total area of vertical stirrup reinforcement required is

………………………………………………….5.97(R1) If foyc = 1.4 fyc for grade 60 rebar design , placing this amount vertical

reinforcement can be difficult. Horizontal hoops are needed, that amount of hoop reinforcement is given by

……………………………………………………5.99(R1) The mechanism (c) required amount of vertical beam stirrups reinforcement is

…………………………….………………………5.100(R1) And the additional area of beam bottom reinforcement required is thus :

………………………………………………….5.101(R1)

Ajv 0.25Ascfoycfyv

⋅:=

ρ s0.6Asc foyc⋅

la2 fyh⋅

:=

Ajv 0.125Ascfoycfyv

⋅:=

ΔAsb 0.0625Ascfoycfyb

⋅:=

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SUBSTRUCTURE


P3 Expansion Coping

Page 1129

Detailed Design Study ofNorth Java CorridorFlyover Project

BALARAJA FLYOVERCoping Design

Pier P3

Detailed Design - Substructure

KATAHIRA & ENGINEERSINTERNATIONAL

Project: Detailed Design Study ofNorth Java Corridor Flyover Project

Calculation: Detailed Design SubstructureBalaraja FlyoverCoping Design - Pier P3

Layout

Katahira & Engineers InternationalPage 1130



Pier P3


Coping Cross Section

Pier Coping Data

Overall width of deck B 13000 mmOffset to bearing - PC deck h1c 3175 mmOffset to bearing - Steel deck h1s 3075 mmEdge distance h2 800 mmSide slope width h3 316 mmCantilever length h4 2160 mmBeam ledge width h5 975 mmBeam ledge soffit width h6 950 mmWidth of coping bf 3350 mmTotal depth at coping v1 2732 mmBeam ledge height at column v2 1200 mmBeam ledge height at bearing v3 1200 mmUpstand to const. joint v4 300 mmBearing width W 800 mmBearing length L 620 mmColumn Diameter D 1700 mmConcrete Comp Strength fc 30 MPaRebar Yield Strength fy 390 MPaStrength Reduction Factor - Bending 0.8Strength Reduction Factor - Shear 0.7Mod. Elasticity - Concrete Ec 27628 MPaMod. Elasticity - Steel Es 200000 MPaModular Ratio 7.24Height of piers supporting deck H 9912 mmLength of deck btwn. Joints Ld 73.6 mDeck skew Sk 0 Deg

B B mm⋅:=

h1c h1c mm⋅:=

h1s h1s mm⋅:=

h2 h2 mm⋅:=

h3 h3 mm⋅:=

h4 h4 mm⋅:=

h5 h5 mm⋅:=

h6 h6 mm⋅:=

bf bf mm⋅:=

v1 v1 mm⋅:=

v2 v2 mm⋅:=

v3 v3 mm⋅:=

v4 v4 mm⋅:=

D D mm⋅:=

L L mm⋅:=

W W mm⋅:=

fc fc MPa⋅:=

fy fy MPa⋅:=

Ec Ec MPa⋅:=

Es Es MPa⋅:=




Pier P3


Analysis OutputThe analysis output for deck reactions at the expansion piers, obtained from the SAP 3D model, is presented below for:

Nominal deck dead load case - used for erection case•Nominal superimposed dead load case•ULS Combination 1 - live load •SLS Combination 1 - live load•Nominal earthquake effects (R=1.0)•

The frame elements selected are as follows:D34 - end span frame in span 3 (PC deck) adjacent to pier 3•D41 - end span frame in span 3 (Steel Deck) adjacent to pier 3•

NOMINAL DECK DEAD LOAD

Frame Station V2 SHEAR VERT

V3 SHEAR TRANS

T TORSION

Text m KN KN KN-mDECK34 Min 1342.9 -0.4 -0.6DECK34 Max 1342.9 -0.4 -0.6DECK41 Min -1505.0 -1.1 2.2DECK41 Max -1505.0 -1.1 2.2

TABLE: Element Forces - Deck Dead Load Reactions

V2dl V2dl kN⋅:= V3dl V3dl kN⋅:= Tdl Tdl kN⋅ m⋅:=

NOMINAL SUPERIMPOSED DEAD LOAD


V3 SHEAR TRANS

T TORSION

Text m KN KN KN-mDECK34 Min 244.6 0.0 0.0DECK34 Max 244.6 0.0 0.0DECK41 Min -289.9 -0.2 0.4DECK41 Max -289.9 -0.2 0.4

TABLE: Element Forces -Superimposed Load Reactions

V2sdl V2sdl kN⋅:= V3sdl V3sdl kN⋅:= Tsdl Tsdl kN⋅ m⋅:=




Pier P3


ULS COMBINATION 1 - LIVE LOAD

Frame Step Type

V2 SHEAR VERT

V3 SHEAR TRANS

T TORSION

Text Text KN KN KN-mDECK34 Min 1911.7 -39.1 -900.2DECK34 Max 6180.6 36.3 896.3DECK41 Min -6701.7 -68.4 -951.3DECK41 Max -2076.1 59.8 969.1

TABLE: Element Forces - Deck COMB1 ULS Reactions

V2u V2u kN⋅:= V3u V3u kN⋅:= Tu Tu kN⋅ m⋅:=

SLS COMBINATION 1 - LIVE LOAD

Frame Step Type

V2 SHEAR VERT

V3 SHEAR TRANS

T TORSION


TABLE: Element Forces - Deck COMB1 SLS Reactions

V2s V2s kN⋅:= V3s V3s kN⋅:= Ts Ts kN⋅ m⋅:=




Pier P3


NOMINAL EARTHQUAKE LOAD - EQX (R=1)

Frame Step Type

V2 SHEAR VERT

V3 SHEAR TRANS

T TORSION


TABLE: Element Forces - EQX

V2eqx V2eqx kN⋅:= V3eqx V3eqx kN⋅:= Teqx Teqx kN⋅ m⋅:=

NOMINAL EARTHQUAKE LOAD - EQY (R=1)

Frame Step Type

V2 SHEAR VERT

V3 SHEAR TRANS

T TORSION


TABLE: Element Forces - EQY

V2eqy V2eqy kN⋅:= V3eqy V3eqy kN⋅:= Teqy Teqy kN⋅ m⋅:=

MAXIMUM SHEAR FORCE DUE TO PLASTIC HINGING

The maximum shear force that the bearings will be subject to in the tranverse direction is theshear force due to plastic hinging at the base of the column.

The shear force due to plastic hingingat the base of the column VP 2612 kN⋅:=

This shear force should be carried by the transverse shear key on each bearing shelf or carried bythe bearings directly on on bearing shelf. The shear forces given above for earthquake loadingshould be used in the design if they are smaller than the plastic hinging effects.




Pier P3


Minimum Displacement Requirements (AASHTO LRFD Article 4.7.4.4)

Bridge seat widths at expansion bearings without restrainers, STU's or dampers shall eitheraccommodate the greater of the maximum displacement calculated from the seismic analysisor a percentage of the emprical seat width, N, specified below.

The percentage of N applicable to the bridge seismic zone shall be 150%.

The length of the bridge deckto the adjacent expansion joint Ld Ld m⋅:= Ld 73.6 m=

The height of the columns supporting the deck

H H mm⋅:= H 9912 mm=

Skew of the supportmeasured from line normal to span

Sk 0=

The empirical seat width shall be taken as:

Ν 200 0.0017L

mm⋅+ 0.0067

Hmm⋅+⎛⎜

⎝⎞⎟⎠

1 0.000125 Sk2

⋅+⎛⎝

⎞⎠⋅ mm⋅:=

Ν 267 mm=

Check that the seat width provided is greater than 150% of N:

Seat width h5 975 mm=

150% Ν⋅ 401 mm=

SeatWidth "OK" h5 Ν 150⋅ %≥if

"INADEQUATE" otherwise

:=

SeatWidth "OK"=




Pier P3


Critical Sections for Design of Pier Coping

Design for Erection Case

During erection the deck dead loads are supported by the partially constructed pier coping.

Partial construction is required to accommodate the prestressing jacks of the PC deck.

To take account of construction equipment on the deck a construction load of 2kN/m 2 isapplied over the full deck area in addition to the deck dead load.Cross sectional area of coping during erection:

Ace v3 bf⋅ v2 v3−( ) D⋅+ 113mm h5⋅+ v1 v2−( ) bf 2 h5⋅−( )⋅+:= Ace 6.275 m2=

Coping self weight during erection:

wce Ace 24.5⋅kN

m3⋅:= wce 153.7

kNm

=

PC Deck Dead Load Reaction - max at the bearing

V2pc

V2dl22

Tdl2h1c 2⋅

+:= V2pc 671.5 kN=

Steel Deck Dead Load Reaction - max at the bearing

V2st

V2dl32

Tdl3h1s 2⋅

+:= V2st 752.9 kN=




Pier P3


Erection Load Reaction - 20m PC deck - per bearing - assuming 45% total reaction into bearing

V2ERpc 2kN

m2⋅ B⋅ 20m( )⋅ 45⋅ %

12⋅:= V2ERpc 117.0 kN=

Erection Load Reaction - 31m Steel deck - per bearing - assuming 45% total reaction into bearing

V2ERst 2kN

m2⋅ B⋅ 31m( )⋅ 45⋅ %

12⋅:= V2ERst 181.3 kN=

Design for Flexure (AASHTO LRFD Section 5.7)

Bending moment in coping at face of column during erection

Coping selfweight

Mce wce

max h1c h1s,( ) h2+h32

+D2

.8⋅−⎛⎜⎝

⎞⎟⎠

2

2⋅:= Mce 916.5 kN m⋅=

PC deckreaction

Mpc V2pc h1cD2

.8⋅−⎛⎜⎝

⎞⎟⎠

⋅:= Mpc 1675.5 kN m⋅=

Steel deckreaction

Mst V2st h1sD2

.8⋅−⎛⎜⎝

⎞⎟⎠

⋅:= Mst 1803.1 kN m⋅=

Erection loadreaction

MER V2ERpc h1cD .8⋅

2−⎛⎜

⎝⎞⎟⎠

⋅ V2ERst h1sD .8⋅

2−⎛⎜

⎝⎞⎟⎠

⋅+:= MER 726.2 kN m⋅=

Total bending moment in coping at face of column during erection - Ultimate Limit State

MEU Mce Mpc+ Mst+( ) 1.25⋅ MER 1.5⋅+:=

MEU 6583.3 kN m⋅=

Depth of section:

hs v2 v4+:= hs 1500 mm=

Effective depth of coping during erection:

de hs 100mm−( ):= de 1400 mm=

Width of coping at column:

b bf h6 2⋅−:= b 1450 mm=

Strength reduction factor for flexure:

Φ 0.8=

Determine area of reinforcement, Af, in the coping to resist flexure(ref AASHTO LFRD Article 5.7.3.2.3 -see notes on flexural design for rearrangement of terms):

RMEU

Φ b⋅ de2

⋅ fy⋅:= R 0.0074= Μ

0.85 fc⋅

fy:= Μ 0.0654=

ρ Μ 1 12 R⋅Μ

−−⎛⎜⎝

⎞⎟⎠

⋅:= ρ 0.007902=




Pier P3


Af ρ de⋅ b⋅:=

Af 16040.9 mm2=

Using 32mmφ bars gives total number of bars to be distributed across the coping:

nbfAf

804 mm2⋅

:= nbf 20=

Provide 30 No 32mm φ bars in two layers

Af 30 804⋅ mm2⋅:=

Calculate the stress block factor, β1:

β1 β1 0.85←

β1 β1 0.05fc 28MPa−

7.MPa⋅−← fc 28MPa>if

0.65 β1 0.65<if

:=

β1 0.836=

For rectangular sections, the depth of concrete in compression, c, is given by:

cAf fy⋅

0.85 fc⋅ β1⋅ b⋅:= c 304 mm=

Check that the required amount of reinforcement does not exceed the maximum allowed by the code:

cde

0.22=

MaxLimit "OK"cde

0.42≤if

"EXCEEDED" otherwise

:= MaxLimit "OK"=

Calculate the modulus of rupture, f r ,of the concrete:

fr 0.63fc

MPa⋅ MPa⋅:= fr 3.451 MPa=

Section modulus for the extreme fiber of the section, assuming rectangular section of width b:

Scb hs

2⋅

6:= Sc 0.544 m3

=

Moment resisting by reinforcemen provided

MEUR Φ c⋅ b⋅ dec2

−⎛⎜⎝

⎞⎟⎠

⋅ 0.85⋅ fc⋅:=

MEUR 11236.1 kN m⋅=

The cracking moment, Mcr ,is the given by:

Mcr Sc fr⋅:= Mcr 1876 kN m⋅=




Pier P3


Check that the reinforcement can develop a resistance moment Mr at least equal to the lesser of:

1.2 times the cracking moment Mcr•1.33 times the factored moment required by the applicable strength load combination•

MinimumSteel MrMEURΦ

←

M min 1.2Mcr 1.33 MEU⋅,( )←

"OK" Mr M≥if

"NOT SATISFIED" otherwise

:= MinimumSteel "OK"=

Design for Shear (AASHTO LRFD Section 5.8)

Critical section for shear shall be taken as dv from the internal face of the support (AASHTOLRFD Article 5.8.3.2)

dv 0.9 de⋅ 0.9 de⋅ 0.72 hs⋅>if

0.72 hs⋅ otherwise

:=

dv 1260 mm=

Shear in coping at critical section during erection

Coping self weight Vce wce max h1c h1s,( ) h2+h32

+D2

.8⋅− dv−⎛⎜⎝

⎞⎟⎠

⋅:= Vce 337.1 kN=

PC deck reaction Vpc V2pc:= Vpc 671.5 kN=

Steel deck reaction Vst V2st:= Vst 752.9 kN=

Erection load reaction VER V2ERpc V2ERst+:= VER 298.4 kN=

Total shear force in coping at face of column during erection - Ultimate Limit State

VEU Vce Vpc+ Vst+( ) 1.3⋅ VER 1.3⋅+:=

VEU 2677.9 kN=

Calculate nominal shear resistance of concrete section assuming beam section:

Vc 0.166fc

MPa⋅ b⋅ dv⋅ MPa⋅:=

Vc 1661 kN=

Strength reduction factor for shear:

Φs 0.7=

Required nominal shear resistance of transverse reinforcement:

Vs VsVEUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:= Vs 2164 kN=




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Check that total required nominal shear resistance does not exceed limit:

Vnlimit 0.25 fc⋅ b⋅ dv⋅:= Vnlimit 13702.5 kN=

CHECK "OK" Vc Vs+ Vnlimit≤if

"FAIL" otherwise

:= CHECK "OK"=

Provide 19mmφ shear links wth 4 legs across the section:

φlink 19mm:=

Av πφlink

2

44⋅:= Av 1134 mm2

=

Determine required spacing of transverse reinforcement

st st1Av fy⋅

0.083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if

st2 Vs 0kN>( ) st2 st1≤( )⋅if

st1 otherwise

:= st 257 mm=

Calculate shear stress in order to determine maximum spacing of transverse reinforcement toresist shear:

vuVEU

Φs b dv⋅( )⋅:= vu 2.094 MPa=

Calculate spacing taking into account maximum spacing requirements:

smax min 600mm 0.8dv,( ) vu 0.125fc<if

min 300mm 0.4dv,( ) otherwise

:=

smax 600 mm=

Determine maximum required spacing of transverse reinforcement:

st.erection smax smax st≤if

st otherwise

:=

st.erection 257 mm=




Pier P3


Design for Permanent Condition

Cross sectional area of coping at support:

Ac v3 bf⋅ v2 v3−( ) D⋅+ bf h5 2⋅−( ) v1 v2−( )⋅+ 113mm h5⋅+:= Ac 6.275 m2=

Coping self weight - main body:

wc Ac 24.5⋅kN

m3⋅:= wc 153.7

kNm

=

Coping self weight - cantilever wings:

wcw0.25m 0.45m+

2bf h5 2⋅−( )⋅ 24.5

kN

m3⋅:= wcw 12.0

kNm

=

Superimposed dead load:

wsdl 0.125m bf h5 2⋅−( )⋅ 22.5kN

m3⋅:= wsdl 3.9

kNm

=

Railing dead load - each side:

Wr0.433

2m2 bf h5 2⋅−( )⋅ 24.5

kN

m3⋅:= Wr 7.4 kN=

D live loading on coping:

wd 9.0kN

m2bf h5 2⋅−( )⋅:= wd 12.6

kNm

=

PC Deck Dead Load Reaction - max at the bearing - ULS Comb1

V2upc

V2u2

2

Tu2

h1c 2⋅+:= V2upc 3231.4 kN=

Steel Deck Dead Load Reaction - max at the bearing - ULS Comb1

V2ust

V2u3

2

Tu3

h1s 2⋅+:= V2ust 3505.5 kN=




Pier P3



Nominal bending moment in coping at face of column

Coping main bodyself weight Mcb wc

max h1c h1s,( ) h2+h32

+D2

.8⋅−⎛⎜⎝

⎞⎟⎠

2

2⋅:= Mcb 916.5 kN m⋅=

Coping cantileverself weight Mcw wcw h4⋅ max h1c h1s,( ) h2+ h3+

h42

+D2

.8⋅−⎛⎜⎝

⎞⎟⎠

⋅:= Mcw 121.6 kN m⋅=

Railingweight

Mr Wr max h1c h1s,( ) h2+ h3+ h4+D2

.8⋅−⎛⎜⎝

⎞⎟⎠

⋅:= Mr 42.9 kN m⋅=

Superimposeddead load oncoping

Msdl wsdl

5.75mD .8⋅

2−⎛⎜

⎝⎞⎟⎠

2

2⋅:= Msdl 50.6 kN m⋅=

D live load oncoping

Md wd

5.75mD .8⋅

2−⎛⎜

⎝⎞⎟⎠

2

2⋅:= Md 161.9 kN m⋅=

Ultimate bending moment in coping from max bearing reactions at face of column:

Mupc V2upc h1cD2

.8⋅−⎛⎜⎝

⎞⎟⎠

⋅:= Mupc 8062.4 kN m⋅=PC deckreaction

Steel deckreaction

Must V2ust h1sD2

.8⋅−⎛⎜⎝

⎞⎟⎠

⋅:= Must 8395.7 kN m⋅=

Total bending moment in coping at face of column - Ultimate Limit State

Moment from loads on coping body

MCU Mcb Mcw+ Mr+( ) 1.3⋅ Msdl 2.0⋅+ Md 1.8⋅+:=

MCU 1798.0 kN m⋅=

Moment from max loads on bearings

MBU Mupc Must+:=

MBU 16458.1 kN m⋅=

Total ULS moment MU MCU MBU+:=

MU 18256.1 kN m⋅=

Depth of section:hs v1:=

Effective depth of coping:

de hs 150mm−( ):= de 2582 mm=

Width of coping at column:

b bf h6 2⋅−:=




Pier P3


Strength reduction factor forflexure:

Φ 0.8=

Determine area of reinforcement, Af, in the coping to resist flexure(ref AASHTO LFRD Article 5.7.3.2.3 - see notes on flexural design for rearrangement ofterms):

RMU

Φ b⋅ de2

⋅ fy⋅:= R 0.0061= Μ

0.85 fc⋅

fy:= Μ 0.0654=

ρ Μ 1 12 R⋅Μ

−−⎛⎜⎝

⎞⎟⎠

⋅:= ρ 0.006363=

Af ρ de⋅ b⋅:=

Af 23821 mm2=


nbfAf

804 mm2⋅

:= nbf 29.6=

Provide 36 No 32mmφ bars

nbars 36:= Af nbars 804⋅ mm2⋅:=


β1 β1 0.85←

β1 β1 0.05fc 28MPa−


0.65 β1 0.65<if

:=

β1 0.836=


cAf fy⋅

0.85 fc⋅ β1⋅ b⋅:= c 365 mm=


cde

0.14=

MaxLimit "OK"cde

0.42≤if


:= MaxLimit "OK"=


fr 0.63fc





Pier P3



Scb hs

2⋅

6:= Sc 1.804 m3

=

Moment resisting by reinforcemen provided

MEUR Φ c⋅ b⋅ dec2

−⎛⎜⎝

⎞⎟⎠

⋅ 0.85⋅ fc⋅:=

MEUR 25926.8 kN m⋅=




1.2 times the cracking momen Mcr•1.33 times the factored moment required by the applicable strength load combination•

MinimumSteel MrMEURΦ

←

M min 1.2Mcr 1.33 MEU⋅,( )←

"OK" Mr M≥if



FINAL LAYOUT OF LONGITUDINAL REBAR




Pier P3


CHECK STRESSES AT SLS (AASHTO LRFD Article 5.7.3.4)

With reference to the illustration above - the final layout of reinforcement features three layers ofrebar as follows:

Layer 1 - 36 No. 32mmφ bars Af1 36 804⋅ mm2:= d1 v1 150mm−:=

Af1 28944 mm2= d1 2582 mm=

Layer 2 - 28 No. 32mmφ bars Af2 26 490⋅ mm2:= d2 v2 v4+ 100mm−:=

Af2 12740 mm2= d2 1400 mm=

Layer 3 - 20 No. 25mmφ bars Af3 20 491⋅ mm2:= d3 v2:=

Af3 9820 mm2= d3 1200 mm=

Total area AT Af1 Af2+ Af3+:=

AT 51504 mm2=

Lever arm factor B Af1 d1⋅ Af2 d2⋅+ Af3 d3⋅+:=

The depth of concrete in compression, C, at the serviceability limit state, assuming threelevels of rebar, is given by:

CAT

2 2 b⋅ B⋅α

+ AT−

bα⋅:= C 796 mm=

PC Deck Dead Load Reaction - max at the bearing - SLS Comb1

V2spc

V2s2

2

Ts2

h1c 2⋅+:= V2spc 2065.7 kN=

Steel Deck Dead Load Reaction - max at the bearing - SLS Comb1

V2sst

V2s3

2

Ts3

h1s 2⋅+:= V2sst 2317.7 kN=

Service bending moment in coping from max bearing reactions at face of column:

PC deck reaction Mspc V2spc h1cD2

.8⋅−⎛⎜⎝

⎞⎟⎠

⋅:= Mspc 5154.0 kN m⋅=

Steel deck reaction Msst V2sst h1sD2

.8⋅−⎛⎜⎝

⎞⎟⎠

⋅:= Msst 5550.9 kN m⋅=

Total bending moment in coping at face of column - Service Limit State

Moment from loads on coping body

MCS Mcb Mcw+ Mr+( ) 1.0⋅ Msdl 1.0⋅+ Md 1.0⋅+:=

MCS 1293.6 kN m⋅=

Moment from max loads on bearings

MBS Mspc Msst+:= MBS 10704.9 kN m⋅=




Pier P3


Total SLSmoment

MS MCS MBS+ 0.5 Mce Mpc+ Mst+( )⋅−:=

MS 9800.9 kN m⋅=

NOTE!: The maximum service moment has been reduced to account for the staged constructionof the coping. 50% of the erection stage moments - carried by the lower reinforcement - havebeen subtracted. The full amount of the erection stage moments have not been subtrated giventhat creep effects will transfer loads to the permanent stage.

Calculate the maximum stress in the reinforcement (IN LAYER 1) at SLS:

fsMS

Af1 d1C3

−⎛⎜⎝

⎞⎟⎠

⋅ Af2d2 C−

d1 C−⋅ d2

C3

−⎛⎜⎝

⎞⎟⎠

⋅+ Af3d3 C−

d1 C−⋅ d3

C3

−⎛⎜⎝

⎞⎟⎠

⋅+

:=

fs 132 MPa=

Check that stress in reinforcement does not exceed limit, fsa:

Crack width parameter Z 30000N

mm⋅:=

Depth of concrete fromextreme tensile fiber to center of bar dc 150mm:=

Area of concrete with same centoid per bar Abf 2 h5⋅−( ) dc⋅ 2⋅

nbars:=

fsa fsaZ

dc A⋅( )1

3

←

fsa fsa 170MPa<if

170 MPa⋅ otherwise

:=

fsa 170 MPa=

StressCheck "OK" fs fsa≤if


:= StressCheck "OK"=

Calculate forces in section to check calculation result:

Total force in rebar

TS Af1 Af2d2 C−

d1 C−⋅+ Af3

d3 C−

d1 C−⋅+

⎛⎜⎝

⎞⎟⎠

fs⋅:= TS 4697 kN=

Stress in concrete

fsc fsC

d1 C−⋅

1α⋅:= fsc 8.1 MPa=

Force in concrete

CC fscb C⋅2

⋅:= CC 4697 kN=




Pier P3




dv 0.9 de⋅ 0.9 de⋅ 0.72 hs⋅>if


:=

dv 2323.8 mm=

Shear in coping at critical section due to loads applied on coping body

Coping self weight Vc wc max h1c h1s,( ) h2+h32

+D2

.8⋅− dv−⎛⎜⎝

⎞⎟⎠

⋅:= Vc 173.6 kN=

Coping cantileverself weight

Vcw wcw h4⋅:= Vcw 25.9 kN=

Railingweight

Vr Wr:= Vr 7.4 kN=

Superimposeddead load on coping

Vsdl wsdl 5.75mD .8⋅

2− dv−⎛⎜

⎝⎞⎟⎠

⋅:= Vsdl 10.8 kN=

D live load oncoping

Vd wd 5.75mD .8⋅

2− dv−⎛⎜

⎝⎞⎟⎠

⋅:= Vd 34.6 kN=

Ultimate shear force in coping from max bearing reactions at face of column:

Vpc V2upc:= Vpc 3231.4 kN=PC deck reaction

Vst V2ust:= Vst 3505.5 kN=Steel deck reaction

Total shear force in coping at face of column - Ultimate Limit State

Shear from loadson coping body

VCU Vc Vcw+ Vr+( ) 1.3⋅ Vsdl 2.0⋅+ Vd( ) 1.8⋅+:=

VCU 353.0 kN=

Shear from max loads on bearings

VBU Vpc Vst+:=

VBU 6736.9 kN=

Total shear force VU VCU VBU+:=

VU 7090 kN=


Vc 0.166fc


Vc 3064 kN=


Φs 0.7=




Pier P3



Vs VsVUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:=

Vs 7065 kN=




"FAIL" otherwise

:= CHECK "OK"=

Provide 19mmφ shear links wth 4 legs across the section

φlink 19mm:=

Av πφlink

2

44⋅:= Av 1134 mm2

=


st st1Av fy⋅

0.083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st 145 mm=


vuVU

Φs b dv⋅( )⋅:= vu 3.006 MPa=




:=

smax 600 mm=


st smax smax st≤if

st otherwise

:= st 145 mm=

Provide 19mmφ links at 100c/c in outer legs and 19mmφ links at 200c/c in inner legs -giving 150mm c/c spacing overall




Pier P3


Design of Cantilever Slab

Equivalent Width of Deck Overhang (AASHTO LRFD Table 4.6.2.1.3-1)

Width of coping slab at support:

b bf 2 h5⋅−:= b 1400 mm=

Distance from outermost load to point of support:

X h4 300mm− 350mm−( ):= X 1510 mm=

Equivalent width of deck overhang:

bew bew 570mm 0.416 X⋅+←

bew bew b≤if

b otherwise

:= bew 1198 mm=

Check Deflection of Cantilever Slab (AASHTO LRFD Article 2.5.2.6.2)

Depth of cantilever slab in main deck section at support:

hds 450mm:=

Span length of cantilever slab in main deck section:

lds 2645mm:=

Depth of deck overhang at support to match main deck outline:

hcs1 hds 250mm−( ) h4lds⋅ 250mm+:= hcs1 413 mm=

Depth of deck overhang at outermost load point:

hcs2 hds 250mm−( ) 350mm 300mm+

lds⋅ 250mm+:= hcs2 299 mm=




Pier P3


Depth of deck overhang at innermost load point:

hcs3 hcs3 hds 250mm−( ) 350mm 300mm+ 1750mm+

lds⋅ 250mm+←

hcs3 hcs3 hcs1≤if

hcs1 otherwise

:=

hcs3 413 mm=

Distance from innermost load to point of support:

X3 X3 h4 300mm− 350mm− 1750mm−←

X3 X3 0>if

0m otherwise

:= X3 0 mm=

Moment of inertia of equivalent width of deck overhang assuming cracked section:

at support Iew1 40 %⋅bew hcs1

3⋅

12⋅:=

at outer load point Iew2 40 %⋅bew hcs2

3⋅

12⋅:=

at innermost load point Iew3 40 %⋅bew hcs3

3⋅

12⋅:=

The deflection under the applied wheel loads is then given by (increaseed by 30% to account fordynamic loading):

δT112.5 kN⋅ 130⋅ %

Ec

0

X

xx2

Iew2 Iew1 Iew2−( ) xX⋅+

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦

⌠⎮⎮⎮⎮⌡

d

0

X3

xx2

Iew3 Iew1 Iew3−( ) xX3⋅+

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦

⌠⎮⎮⎮⎮⌡

d+

⎡⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎦

⋅:=

δT 2.61 mm=

Check that the deflection does not exceed the limit for vehicular load on cantilever arms:

δLIMITX

300:= δLIMIT 5.03 mm=

DEFLECTIONCHECK "OK" δT δLIMIT≤if

"FAIL" otherwise

:=

DEFLECTIONCHECK "OK"=


Nominal bending moment in slab atsupport


Mcw wcwh42

2⋅:= Mcw 28.0 kN m⋅=




Pier P3


Railingweight

Mr Wr h4 0.15m+ 0.25m−( )⋅:= Mr 15.3 kN m⋅=

Superimposed deadload on coping

Msdl wsdlh4 350mm−( )2

2⋅:= Msdl 6.4 kN m⋅=

T live load oncoping

Mt 112.5 kN⋅ X X3+( )⋅:= Mt 169.9 kN m⋅=

Total bending moment in slab at face of support - Ultimate Limit State

MSU Mcw Mr+( ) 1.3⋅ Msdl 2.0⋅+ Mt 1.3⋅( ) 1.8⋅+:=

MSU 466.7 kN m⋅=

Depth of section:

hcs hcs1:= hcs 413 mm=

Effective depth of slab:

de hcs 90mm−( ):= de 323 mm=

Effective width of coping slab at support:

bew 1198 mm=

Strength reduction factor for flexure: Φ 0.8=

Determine area of reinforcement, Af, in the coping to resist flexure(ref AASHTO LFRD Article 5.7.3.2.3 - see notes on flexural design for rearrangement of terms):

RMSU

Φ bew⋅ de2

⋅ fy⋅:= R 0.0119= Μ

0.85 fc⋅

fy:= Μ 0.0654=

ρ Μ 1 12 R⋅Μ

−−⎛⎜⎝

⎞⎟⎠

⋅:= ρ 0.013294=

Af ρ de⋅ b⋅:= b 1400 mm=

Af 6017.5 mm2=


nbfAf

804 mm2⋅

:= nbf 7.5=


nbars 10:= Af nbars 804⋅ mm2⋅:= Af 8040 mm2

=





Pier P3


β1 β1 0.85←

β1 β1 0.05fc 28MPa−


0.65 β1 0.65<if

:=

β1 0.836=


cAf fy⋅

0.85 fc⋅ β1⋅ b⋅:= c 105 mm=


cde

0.33=

MaxLimit "OK"cde

0.42≤if


:= MaxLimit "OK"=


fr 0.63fc



Scb hcs

2⋅

6:= Sc 0.04 m3

=





MinimumSteel MrMSUΦ

←

M min 1.2Mcr 1.33 MEU⋅,( )←

"OK" Mr M≥if






Pier P3



The depth of concrete in compression, C, at the serviceability limit state, assuming one levelof rebar, is given by:

CAf

2 2 b⋅ Af⋅ de⋅

α+ Af−

bα⋅:= C 128 mm=

Calculate lever arm, z:

z deC3

−:= z 281 mm=

Total bending moment in slab at face of support - Service Limit State

MSS Mcw Mr+( ) 1.0⋅ Msdl 1.0⋅+ Mt 1.3⋅( ) 1.0⋅+:=

MSS 270.6 kN m⋅=

Calculate the maximum stress in the reinforcement at SLS:

fsMSSAf z⋅

:= fs 120 MPa=



mm⋅:=


Area of concrete with same centoidper bar A

b dc⋅ 2⋅

nbars:=

fsa fsaZ

dc A⋅( )1

3

←

fsa fsa 170MPa<if


:=

fsa 170 MPa=





Total force in rebar Stress in concrete

TS Af( ) fs⋅:= TS 964 kN= fsc fsC

de C−⋅

1α⋅:= fsc 10.8 MPa=

Force in concrete

CC fscb C⋅2

⋅:= CC 964 kN=




Pier P3




dv 0.9 de⋅ 0.9 de⋅ 0.72 hcs⋅>if

0.72 hcs⋅ otherwise

:=

dv 298 mm=

Shear in coping slab at critical section due to loads applied on coping body


Vcw wcw h4 dv−( )⋅:= Vcw 22.4 kN=

Railing weight Vr Wr:= Vr 7.4 kN=


Vsdl wsdl h4 600mm− dv−( )⋅:= Vsdl 5.0 kN=


Vt 112.5 kN⋅:= Vt 112.5 kN=

Total shear force in coping slab at face of support - Ultimate Limit State


VSU Vcw Vr+( ) 1.3⋅ Vsdl 2.0⋅+ Vt 1.4⋅( ) 1.8⋅+:=

VSU 332.2 kN=


Vc 0.166fc


Vc 379 kN=


Φs 0.7=


Vs VsVSUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:= Vs 96 kN=




"FAIL" otherwise

:= CHECK "OK"=




Pier P3


Provide 13mmφ shear links with 4 legs across the section

φlink 13mm:=

Av πφlink

2

44⋅:= Av 531 mm2

=


st st1Av fy⋅

0.083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st 644 mm=


vuVSU

Φs b dv⋅( )⋅:= vu 1.139 MPa=




:=

smax 238 mm=



st otherwise

:=

st 238 mm=

Provide 13mmφ links at 150c/c




Pier P3


Beam Ledge Design AASHTO LRFD Article 5.13.2.5

General, Article 5.13.2.5.1

As illustrated below, beam ledges shall resist:

Flexure, shear and horizontal forces at the loaction of Crack 1;•Tension force in the supporting element at the location of Crack 2;•Punching shear at points of loading at the location of Crack 3; and•Bearing forces at the location of Crack 4•




Pier P3


Design for Shear, Article 5.13.2.5.2

Design of beam ledges for shear shall be in accordance with the requirements of shear friction inArticle 5.8.4.

The width of the concrete face assumed to participate shall not exceed the width ilustrated below:

Edge distance of bearing C h2:= C 800 mm=

Depth of beam ledge h v3:= h 1200 mm=

Effective depth de h 65mm−:= de 1135 mm=

Width of the interface bv 2 C⋅:= bv 1600 mm=

Area of concreteresisting shear transfer

Acv bv de⋅:= Acv 1.816 m2=

Loads on Ledge

Ultimate shear force in coping from max bearing reactions at face ofcolumn:

PC deckreaction

Vupc V2upc:= Vupc 3231.4 kN=

Steel deckreaction

Vust V2ust:= Vust 3505.5 kN=

Maximum design shearforce

Vu max Vupc Vust,( ):= Vu 3506 kN=




Pier P3


Calculate limiting nominal shear strength

Vnlimit 0.2 fc⋅ Acv⋅ 0.2 fc⋅ Acv⋅ 5.5Acv

mm2⋅ N⋅<if

5.5Acv

mm2⋅ N⋅ otherwise

:=

Vnlimit 9988 kN=

Calculate required nominal shear strength and check beam ledge depth

VnVuΦs

:= Vn 5007.9 kN=

BeamLedge "OK" Vn Vnlimit≤if


:=BeamLedge "OK"=

Calculate shear friction reinforcement, Avf

Cohesion and friction values for monolithically cast concrete

c 1.0MPa:=

λ 1.00:=

μ 1.4 λ⋅:=

Shear reinforcement is then given by:

Avf Avf1Vn c Acv⋅−

fy μ⋅←

Avf2 0.35bvmm⋅

MPafy

⋅ mm2⋅←

Avf Avf1← Avf1 Avf2>if


0VnAcv

0.7MPa<if

:=

VnAcv

2.758 MPa=

Avf 5845.9 mm2=




Pier P3


Design for Flexure and Horizontal Force, Article 5.13.2.5.3

The area of total primary tension reinforcement shall satisfy the requirements of Article 5.13.2.4.2.

The primary tension reinforcement shall be spaced uniformly with the region 2C.

The section at the face of the support shall be designed to resist simultaneously a factored shearforce Vu, a factored moment Mu and a concurrent factored horizontal tensile force Nuc.

Nuc shall not be taken to be less than 0.2Vu and shall be regarded as a live load, even where itresults from creep, shrinkage or temperature change.

These provisions apply to beam ledges:

• with a shear span-to-depth ratio av/de not greater than unity

• subject to a horizontal tensile force Nuc not larger than Vu

The depth at outside edge of bearing shall not be less than 0.5de, where d e is effective depth.

Horizontal tensile force Nuc 0.2 Vu⋅:= Nuc 701.1 kN=

Shear span avh52

:= av 488 mm=

Design Moment Mu Vu av⋅ Nuc h de−( )⋅+:= Mu 1754.5 kN m⋅=

Design the primary tensile force reinforcement As:

Strength reduction factor for bending Φ 0.8=

Width of section b 2 C⋅:= b 1600 mm=

Determine area of primary reinforcement, As, to resist flexure(ref AASHTO LFRD Article 5.7.3.2.3 - see notes on flexural design for rearrangement of terms):

RMu

Φ b⋅ de2

⋅ fy⋅:= R 0.0027= Μ

0.85 fc⋅

fy:= Μ 0.0654=

ρ Μ 1 12 R⋅Μ

−−⎛⎜⎝

⎞⎟⎠

⋅:= ρ 0.002788=

As ρ b⋅ de⋅:= As 5062.5 mm2=

Design the tensile force reinforcement An:

AnNucΦ fy⋅

:= An 2247 mm2=




Pier P3


Check that the Area of Primary Reinforcement, As, satisfies code requirements:

As As As23

Avf⋅ An+>if

23

Avf⋅ An+ otherwise

:=

As 0.04fcfy⋅ b⋅ de⋅ As 0.04

fcfy⋅ b⋅ de⋅<if

As otherwise

:=

b 1.6 m=

As 6144 mm2= As

201mm230.569=

AsAsb

:= As 3.84mm2

mm=

check area of steel required

ρREQUIREDAs m⋅

b de⋅100⋅:= ρREQUIRED 0.21= PERCENT

Determine area of closed stirrups or ties, Ah:

Ah 0.5 As b⋅ An−( )⋅:= Ah 1949 mm2=

This reinforcement shall be uniformly distributed within two-thirds of the effective depth of thebeam ledge adjacent to As.

Anchorage of primary reinforcement:

At the front face of the beam ledge, the primary tension reinforcement, As, shall be anchoredby one of the following:

a) a structural weld to a transverse bar of at least equal size; weld to be designed to developspecified yield strength fy of As bars

b) bending primary tension bars As back to form a horizontal loop

c) some other means of positive anchorage

The bearing area of load on the bracket or corbel shall not project beyond interior face of transverseanchor bar (if one is provided).




Pier P3


Design for Punching Shear, Article 5.13.2.5.4

The truncated pyramids assumed as failure surfaces for punching shear, as illustrated below,shall not overlap.

Applied ultimate reaction Vu 3506 kN=

Width of bearing L 620 mm=

Length of bearing W 800 mm=

Effective depth de 1135 mm=

Bearing pad spacing S min h1c h1s,( ) 2⋅:= S 6150 mm=

Nominal punching shear resistance, Vn, shall be taken as:At interior pads, or exterior pads, where the end distance C is greater than half the pad spacing•S/2:

Vn1 0.328fc

MPa⋅ MPa⋅ W 2 L⋅+ 2 de⋅+( )⋅ de⋅:= Vn1 8788 kN=

At exterior pads where the end distance C is less than half the pad spacing S/2 and C-0.5W•is less than de:

Vn2 0.328fc

MPa⋅ MPa⋅ W L+ de+( )⋅ de⋅:= Vn2 5210 kN=

At exterior pads where the end distance C is less than half the pad spacing S/2 but C-0.5W•is greater than de:

Vn3 0.328fc

MPa⋅ MPa⋅ 0.5W L+ de+ C+( )⋅ de⋅:= Vn3 6025 kN=




Pier P3


Determine the nominal punching resistance:

Vn Vn1 CS2

≥if

Vn2 CS2

<⎛⎜⎝

⎞⎟⎠

C 0.5 W⋅−( ) de≤⎡⎣ ⎤⎦⋅if

Vn3 otherwise

:=

Vn 5210 kN=

Check that the nominal punching resistance is adequate:

PunchingShearCHECK "SATISFIED" Vn Φs⋅ Vu≥if

"FAIL" otherwise

:=

PunchingShearCHECK "SATISFIED"=




Pier P3


Design for Hanger Reinforcement, Article 5.13.2.5.5

Hanger reinforcement specified herein shall be provided in addition to the lesser shearreinforcement required on either side of the beam reaction being supported.

The distance from the top of the ledge to the compression reinforcement as illustrated above is d f.

bf 3350 mm= df de 40mm−:= df 1095 mm=

The nominal shear resistance of ledges of inverted T-beams shall be the lessor of the following:

VN1 Ahr s,( )Ahr fy⋅

sS⋅:= (Equation 1)

(Equation 2)VN2 Ahr s,( ) 0.165

fcMPa

⋅ MPa⋅ bf⋅ df⋅Ahr fy⋅

sW 2df+( )⋅+:=

In the case of the applied design, the edge distance between the exterior bearing pad and the endof the shelf is less than df. Equation 2 above is therefore modified as shown below:

(Equation 3)VN2 Ahr s,( ) 0.165

fcMPa

⋅ MPa⋅ bf⋅ df⋅Ahr fy⋅

s2 C⋅( )⋅+:=

Try 25mm φ rebar at 100mm c/c

Area of one leg of hanger reinforcement Ahr and spacing s are then:

Ahrπ 25mm( )2⋅

4:= s 100mm:=

Total number of bars required:

nbars2 C⋅

s:= nbars 16.0=

This gives:

VN1 Ahr s,( ) 11774 kN=

VN2 Ahr s,( ) 6378 kN=

The minimum nominal shear resistance is then given by:

Vn min VN1 Ahr s,( ) VN2 Ahr s,( ),( ):=

Vn 6378 kN=




Pier P3


Max shear force from the bearings:

Vu 3506 kN=

Check that the nominal shear resistance is adequate

HangerCHECK "SATISFIED" Vn Φs⋅ Vu≥if

"FAIL" otherwise

:=

HangerCHECK "SATISFIED"=

Check Torsional Requirements (AASHTO LRFD Section 5.8)

Check the torsional moment requirements of the beam ledge assuming the bearings (on the steeldeck side) are fully loaded and the opposite bearings (on the PC deck side) are loaded only withpermanent load:


V2pc 672 kN=

Superimposed Dead Load PC Deck Reaction - max at the bearing

V2sdl2245 kN=

Ultimate limit state reaction in bearing under permanent load:

Vupc V2pc 1.3⋅ V2sdl22.0⋅+:=

Vupc 1362 kN=

Calculate torsion in the coping:

Tc Vu Vupc−( )bf2

h52

−⎛⎜⎝

⎞⎟⎠

:=

Tc 2545 kN m⋅=

Associated shear force:

VTU VCU Vust+ Vupc+:=

VTU 5221 kN=

The coping will resist torsion moment with two torson blocks as illustrated below. The torsionmoment will be distributed into each torsion block in accordance with the relative area of eachblock.




Pier P3


Dimensions of the torsion blocks are as follows:

Torsion block 1 h1 v1:= h1 2732 mm=

b1 bf h5 2⋅−:= b1 1400 mm=

Torsion block 1 h223

v3⋅:= h2 800 mm=

b2 bf:= b2 3350 mm=

Area enclosed with centerline of transverse rebar

Torsion block 1 Aoh1 b1 40mm 2⋅− 19mm−( ) h1 40mm 2⋅− 19mm−( )⋅:=


Area enclosed by shear flow path

Torsion block 1 Ao1 0.85Aoh1:=


Calculate torsion moments carried by each block:

Torsion block 1 Tc1Ao1

Ao1 Ao2+Tc⋅:= Tc1 1525 kN m⋅=


Ao1 Ao2+Tc⋅:= Tc2 1020 kN m⋅=




Pier P3


Transverse Reinforcement for Shear and Torsion - Torsion Block 1

Depth of section:

hs v1:=


de hs 150mm−( ):= de 2582 mm=

Width of coping:

b bf h5 2⋅−:= b 1400 mm=


dv 0.9 de⋅ 0.9 de⋅ 0.72 hs⋅>if


:=

dv 2323.8 mm=

Calculate nominal shear resistance of concrete section assuming beamsection:

Vc 0.166fc


Vc 2958 kN=


Φs 0.7=


Vs VsVTUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:=

Vs 4500 kN=


φlink 19mm:=

Av πφlink

2

44⋅:= Av 1134 mm2

=


st1 st1Av fy⋅

0.083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st1 228 mm=




Pier P3


For reinforced concrete the angle of inclination of diagonal stresses, θ, can be assumed tobe 45 degrees:

θ 45deg:=

Determine required transverse reinforcement fortorsion:

Area of one leg of transversetorsion reinforcement At π

φlink2

4:= At 284 mm2

= φlink 19 mm=

Required spacing of torsional reinforcement st2

2 Ao1⋅ At⋅ fy⋅ cot θ( )⋅

Tc1Φs⋅:= st2 296 mm=

Calculate combined spacing of shear and torsion transversereinforcement:

1st1

1st2

+⎛⎜⎝

⎞⎟⎠

1−128.833 mm=

Provide 19mm dia transverse reinforcement at 100mm c/c in main body of coping



φlink 16mm:=

Av πφlink

2

42⋅:= Av 402 mm2

=

Determine required transverse reinforcement fortorsion:


φlink2

4:= At 201 mm2

= φlink 16 mm=

Required spacing of torsionalreinforcement

st22 Ao2⋅ At⋅ fy⋅ cot θ( )⋅


Provide 16mm dia transverse reinforcement at 200mm c/c in beam ledges




Pier P3


Design for Loads from Deck Jacking

AASHTO LRFD requires that the beam ledge is designed to resist deck jacking forces.

The deck jacking loads shall not be less than 1.3 times the permanent load reaction at the bearingadjacent to the point of jacking.


Vdlpc

V2dl22

Tdl2h1c 2⋅

+:= Vdlpc 671.5 kN=


Vdlst

V2dl32

Tdl3h1s 2⋅

+:= Vdlst 752.9 kN=

PC Deck Superimposed Dead Load Reaction - max at the bearing

Vsdlpc

V2sdl22

Tsdl2h1c 2⋅

+:= Vsdlpc 122.3 kN=

Steel Deck Superimposed Dead Load Reaction - max at the bearing

Vsdlst

V2sdl32

Tsdl3h1s 2⋅

+:= Vsdlst 145.0 kN=

Maximum design load reaction due to deck jacking:

VJack 1.3 max Vdlpc Vsdlpc+ Vdlst Vsdlst+,( ):=

VJack 1167 kN=

Assuming the worst case for positioning of the jack and ignoring width of loaded area from jack gives:

Horizontal tensile force Nuc 0.2 VJack⋅:= Nuc 233.4 kN=

Shear span av h5:= av 975 mm=

Design Moment Mu VJack av⋅ Nuc h de−( )⋅+:= Mu 815.4 kN m⋅=

Loaded width during jacking w 1000mm:= w 1000 mm=



Width of the interface bv w:= bv 1000 mm=






Pier P3


Calculate required nominal shear strength

VnVJackΦs

:= Vn 1667.5 kN=



c 1.0MPa:=

λ 1.00:=

μ 1.4 λ⋅:=



fy μ⋅←

Avf2 0.35bvmm⋅

MPafy

⋅ mm2⋅←



0VnAcv

0.7MPa<if

:=

VnAcv

1.469 MPa=

Avf 975.2 mm2=



Width of section b 1000mm:=


RMu

Φ b⋅ de2

⋅ fy⋅:= R 0.0020= Μ

0.85 fc⋅

fy:= Μ 0.0654=

ρ Μ 1 12 R⋅Μ

−−⎛⎜⎝

⎞⎟⎠

⋅:= ρ 0.002061=

As ρ b⋅ de⋅:= As 2339.6 mm2=


AnNucΦ fy⋅

:= An 748 mm2=




Pier P3



As As As23

Avf⋅ An+>if

23


:=



As otherwise

:=

As 3492 mm2=

AsAsb

:= As 3492mm2

m=

check area of steelrequired

ρREQUIREDAs m⋅



Ah 0.5 As b⋅ An−( )⋅:= Ah 1372 mm2=


Design for Hanger Reinforcement

The nominal shear resistance of ledges of inverted T-beams shall be the lessor of the following:

VN1 Ahr s,( )Ahr fy⋅

sS⋅:=

VN2 Ahr s,( ) 0.165fc

MPa⋅ MPa⋅ bf⋅ df⋅

Ahr fy⋅

sb⋅+:=

Try 19mm φ rebar at 100mm c/c

Area of one leg of hanger reinforcement Ahr and spacing s are then:

Ahrπ 19mm( )2⋅

4:= s 100mm:=

Total number of bars required:

nbarsbs

:= nbars 10.0=

This gives:

VN1 Ahr s,( ) 6800 kN=

VN2 Ahr s,( ) 4421 kN=




Pier P3


The minimum nominal shear resistance is then given by:

Vn min VN1 Ahr s,( ) VN2 Ahr s,( ),( ):=

Vn 4421 kN=

Max shear force from the jack:

VJack 1167 kN=

Check that the nominal shear resistance is adequate

HangerCHECK "SATISFIED" Vn Φs⋅ VJack≥if

"FAIL" otherwise

:=

HangerCHECK "SATISFIED"=

CONCLUSION - PROVIDE SAME REINFORCEMENT FOR MAIN BEAM LEDGE DESIGN ACROSSENTIRE BEAM LEDGE WIDTH - EXCEPT OVER WIDTH OF COLUMN.

JACKING POINTS SHALL NOT BE CLOSER THAN 500mm FROM FACE OF COLUMN.




Pier P3


Design for Loads from Longitudinal Restrainers, Article 3.10.9.5

This calculation note is intended for the design of the reinforced concrete elements of the piercoping supporting the restrainers.

For the design of the restrainers themselves, inlcuding any local bursting reinforcement required,refer to a separatee calculation.

Restrainers shall be designed for a force calculated as the acceleration coefficient times thepermanent load of the lighter of the two adjoining spans.Acceleration coefficient

A 0.40:=

Permanent load of the lighter of the two spans assuming, conservatively that only 40% of thetotal load appears as a reaction at the pier coping:

PERMLOAD V2pc V2sdl2+⎛

⎝⎞⎠

140%

:= PERMLOAD 2290 kN=

Design load in restrainer

RESTLOAD PERMLOAD A⋅:= RESTLOAD 916 kN=

Required nominal shear strength of pier coping supporting the restrainers:

VnrRESTLOAD

Φs:= Vnr 1309 kN=

Assume conservatively that this load is carried at a single point with a 150mm x 150mm bearingplate located at mid height of the PC deck. Given that the applied load will be supported by acoping width that is at least equal to the shear span, av, of the load - design shear frictionreinforcement in accordance with Article 5.13.2.5.2 and Article 5.8.4.




Pier P3


Shear span of the load av v1 v2−1200mm

2−:= av 932 mm=

The width of the concrete face assumed to participate in resistance to shear is as defined below:

bv S min 2 h1c⋅ 2 h1s⋅,( )←

dw 150mm 4 av⋅+←

dw dw S<if

S otherwise

:= bv 3878 mm=

Depth of support h bf h5 2⋅−:= h 1400 mm=






mm2⋅ N⋅<if

5.5Acv


:= Vnlimit 28474.2 kN=

Check that the required nominal shear strength is less than that provided

PierCopingCapacity "OK" Vnr Vnlimit≤if


:=PierCopingCapacity "OK"=



c 1.0MPa:= λ 1.00:= μ 1.4 λ⋅:=


Avf Avf1Vnr c Acv⋅−

fy μ⋅←

Avf2 0.35bvmm⋅

MPafy

⋅ mm2⋅←



0VnrAcv

0.7MPa<if

:=

VnrAcv

0.253 MPa=

Avf 0 mm2=

Conclude that the concrete section alone has adequate capacity to carry the loads from thelongitudinal restrainers without the need for addiional shear reinforcement.




Pier P3


Design the Pier Column/Coping Connection for Flexure and Shear

The pier column/coping conection will be designed to resist the maximum moments and shearforces applied from live loads and earthquake loads.

Analysis OutputThe analysis output for load effects at the top of the expansion piers, obtained from the SAP 3Dmodel, is presented below.

The half live load case is with live load only occupying the lanes in one carriageway to createmaximum transverse moment at the top of the column.

ULTIMATE FLEXURAL DEMAND AT THE TOP OF THE COLUMN

KN KN-m KN-m KN-mCOMBINATION 1 max -5640.0 742.8 0.0 742.8

Full live Load min -10523.7 -752.6 0.0 -752.6COMBINATION 1 max -5840.4 5159.7 0.0 5159.7

Half Live Load min -8317.3 -5171.0 0.0 -5171.0COMBINATION 5 max -4216.6 390.4 1180.1 1243.0

1.0 EQX+ 0.3 EQY min -4593.0 -393.2 -1180.1 1243.9COMBINATION 5 max -4340.9 920.2 465.2 1031.1

0.3 EQX+ 1.0 EQY min -4468.7 -923.0 -465.2 1033.6

P P− kN⋅:= Md Md2 kN⋅ m⋅:=




Pier P3


Design for Flexure in Column (AASHTO LRFD Section 5.7)

The maximum moment to be used in the fexural design of the connection is given below (maximum ofmoments obtained from global analysis and longitudinal moment generated by loading only one deckspan):

Mmax M1 max Md( )←

M2 Tc 2⋅←

max M1 M2,( )

:= Mmax 5171 kN m⋅=

From inspection above the critical case is due to transverse live load occupying lanes on one side ofthe deck only. The associated axial load with this case, P, therefore will be intermediate between themaximum and minimm axial loads identified above for the load configuration.

PP3 P4+

2:= P 7079 kN=

PCACol has been used to design the reinforcement concrete column.

Results of the PCACol design are as follows:

Use 36 number 32mmφ bars in two bar bundles in a single layer ajaccent to top coping

To aid with reinforcement fixing the bars are placed to give a cover of 60 mm longitudinalbars.This will allow the placing of horizontal shear reinforcement in the faces of the coping beamadjacent to the column rebar. The bar layout is shown below:




Pier P3


Design for Shear in Column (AASHTO LRFD Section 5.8)

Depth of section hs D:= hs 1700 mm=

Width of section b D:= b 1700 mm=

Diameter of circlepassing through centers of longitudinal rebar

Dr hs 2 80⋅ 32+( ) mm⋅−:=

Effective depth dehs2

Drπ

+:=


dv 0.9 de⋅ 0.9 de⋅ 0.72 hs⋅>if


:=

dv 1224 mm=

Maximum shear force (from plastic hinging at base of column)

VP 2612 kN=


Vc 0.166fc


Vc 1892 kN=


Φs 0.7=


Vs VsVPΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:=

Vs 1840 kN=




"FAIL" otherwise

:= CHECK "OK"=

Provide 19mmφ spirals

φlink 19mm:=

Av πφlink

2

42⋅:= Av 567 mm2

=




Pier P3



st st1Av fy⋅

0.083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st 147 mm=


vuVP

Φs b dv⋅( )⋅:= vu 1.793 MPa=




:=

smax 600 mm=



st otherwise

:=

st 147 mm=

Provide 19mmφ spirals at 100mm c/c

Note that although this section will not be subject to plastic hinging, the detailing of thereinforcement will follow the requirements at plastic hinge zones:

maximum spacing 100mm•shear links extended into the support for a distance not less than one half the column•diameter columnthe development length of the longitidinal steel shall be 1.25 times that required for the•full yield of the reinforcing bar




Pier P3


Design for Shear at Connection (AASHTO LRFD Section 5.8)

The applied loading on the coping will be carried by shear forces in the connection generatingtension and compression fields as shown below.

The vertical component of the tension field in the connection will be carried by the main columnreinforcement, extended to the top of the connection.

The horizontal component will be carried by additional reinforcement designed to resist thehorizontal shear force generated at the connection.

TENSION FIELD CARRIED BY VERTICAL SHEAR

LINKS

APPLIED LOAD

DIAGONAL TENSION FIELD CARRIED BY VERTICAL AND HORIZONTAL REBAR AT CONNECTION

TENSION FIELD CARRIED BY MAIN HORIZONTAL REBAR

COMPRESSIVE FORCE TENSILE FORCE

Defining offset "a" to the main rebar in the upper coping beam and offset "b" to cenroid ofcompression zone in the lower coping beam resisting the applied moment:

a 150mm:= b 100mm:=

leve

r arm

za

b

T

C

Lever arm in the copingat the face of the conection

z v1 a− b−:= z 2482 mm=

Calculate the tension T and compression C forces at the connection under maximum appliedmoment:

TMmax

z−:= T 2083− kN= C

Mmaxz

:= C 2083 kN=




Pier P3


Check the assumed depth of concrete incompression:

width of coping bw bf 2 h5⋅−:= bw 1400 mm=

depth of concrete in compression

cC

bw 0.85⋅ fc⋅:= c 58 mm=

Given that the depth of concrete in compression is less than the assumed offset to the centroid,accept the assumed values as conservative.

Depth of section hs D:= hs 1700 mm=

Width of section b D:= b 1700 mm=

Diameter of circlepassing through centers of longitudinal rebar in column

Dr hs 2 80⋅ 32+( ) mm⋅−:=

Effective depth dehs2

Drπ

+:= de 1330 mm=


dv 0.9 de⋅ 0.9 de⋅ 0.72 hs⋅>if


:=

dv 1224 mm=

Maximum shear force (from maximum moment applied at the connection)

VU C:= VU 2083 kN=

Calculate nominal shear resistance of concrete section at the connection (refer ASHTO LRFDArticle 5.10.11.4.3):

Vc 1.0fc


Vc 11397 kN=


Φs 0.7=


Vs VsVUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:=

Vs 0 kN=




"FAIL" otherwise

:= CHECK "OK"=




Pier P3


Provide 19mmφ shear links wth 2 legs across the section - one each face

φlink 19mm:=

Av πφlink

2

42⋅:= Av 567 mm2

=


st st1Av fy⋅

0.083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st 286 mm=

Calculate shear stress in order to determine maximum spacing of transverse reinforcement to resistshear:

vuVP

Φs b dv⋅( )⋅:= vu 1.793 MPa=




:=

smax 600 mm=



st otherwise

:=

st 286 mm=

Provide 19mmφ bar at 200mm c/c each face




SUBSTRUCTURE


P6 Expansion Coping

Page 1181



Pier P6


KATAHIRA & ENGINEERSINTERNATIONAL

Project: Detailed Design Study ofNorth Java Corridor Flyover Project

Calculation: Detailed Design SubstructureBalaraja FlyoverCoping Design - Pier P6

Layout




Pier P6


Coping Cross Section

Pier Coping Data

Overall width of deck B 13000 mmOffset to bearing - PC deck h1c 3175 mmOffset to bearing - Steel deck h1s 3075 mmEdge distance h2 800 mmSide slope width h3 316 mmCantilever length h4 2160 mmBeam ledge width h5 1200 mmBeam ledge soffit width h6 950 mmWidth of coping bf 3800 mmTotal depth at coping v1 2732 mmBeam ledge height at column v2 1200 mmBeam ledge height at bearing v3 1200 mmUpstand to const. joint v4 300 mmBearing width W 800 mmBearing length L 800 mmColumn Diameter D 1100 mmDistance Column to Column dc 6000 mmConcrete Comp Strength fc 30 MPaRebar Yield Strength fy 390 MPaStrength Reduction Factor - Bending 0.8Strength Reduction Factor - Shear 0.7Mod. Elasticity - Concrete Ec 27628 MPaMod. Elasticity - Steel Es 200000 MPaModular Ratio 7.24Height of piers supporting deck H 9912 mmLength of deck btwn. Joints Ld 73.6 mDeck skew Sk 0 Deg

B B mm⋅:=

h1c h1c mm⋅:=

h1s h1s mm⋅:=

h2 h2 mm⋅:=

h3 h3 mm⋅:=

h4 h4 mm⋅:=

h5 h5 mm⋅:=

h6 h6 mm⋅:=

bf bf mm⋅:=

v1 v1 mm⋅:=

v2 v2 mm⋅:=

v3 v3 mm⋅:=

v4 v4 mm⋅:=

D D mm⋅:=

dc dc mm⋅:=

L L mm⋅:=

W W mm⋅:=

H H mm⋅:=

fc fc MPa⋅:=

fy fy MPa⋅:=

Ec Ec MPa⋅:=

Es Es MPa⋅:=




Pier P6


Analysis OutputThe analysis output for deck reactions at the expansion piers, obtained from the SAP 3D model, is presented below for:

Nominal deck dead load case - used for erection case•Nominal superimposed dead load case•ULS Combination 1 - live load •SLS Combination 1 - live load•

Nominal earthquake effects (R=1.0)•The frame elements selected are as follows:•D34 - end span frame in span 3 (PC deck) adjacent to pier 3•D41 - end span frame in span 3 (Steel Deck) adjacent to pier 3•

NOMINAL COPING DEAD LOAD

TABLE: Element Forces - Frames

FrameStatio

n OutputCaseStepTy

pe P V2 M3

Text m Text Text KN KN KN-mC Left 0 DEAD-COPING Max 2.7 459.4 20.6C Left 3 DEAD-COPING Min 2.7 -2.3 -665C Right 0 DEAD-COPING Max 2.6 2.4 -664.9C Right 3 DEAD-COPING Min 2.6 -459.3 20.4

Shear V2 Moment M3

VCop V2 kN⋅:= MCop M3 kN⋅ m⋅:=

NOMINAL DECK DEAD LOAD


V3 SHEAR TRANS

T TORSION

Text m KN KN KN-mDECK64 Min 1504.6 1.8 -4.1DECK64 Max 1504.6 1.8 -4.1DECK71 Min -1331.7 8.5 -0.6DECK71 Max -1331.7 8.5 -0.6

TABLE: Element Forces - Deck Dead Load Reactions

V2dl V2dl kN⋅:= V3dl V3dl kN⋅:= Tdl Tdl kN⋅ m⋅:=

NOMINAL SUPERIMPOSED DEAD LOAD


V3 SHEAR TRANS

T TORSION

Text m KN KN KN-mDECK64 Min 289.5 0.3 -0.7DECK64 Max 289.5 0.3 -0.7DECK71 Min -242.4 1.5 0.0DECK71 Max -242.4 1.5 0.0

TABLE: Element Forces -Superimposed Load Reactions

V2sdl V2sdl kN⋅:= V3sdl V3sdl kN⋅:= Tsdl Tsdl kN⋅ m⋅:=




Pier P6


ULS COMBINATION 1 - LIVE LOAD

Frame Step Type

V2 SHEAR VERT

V3 SHEAR TRANS

T TORSION


TABLE: Element Forces - Deck COMB1 ULS Reactions

V2u V2u kN⋅:= V3u V3u kN⋅:= Tu Tu kN⋅ m⋅:=

SLS COMBINATION 1 - LIVE LOAD

Frame Step Type

V2 SHEAR VERT

V3 SHEAR TRANS

T TORSION


TABLE: Element Forces - Deck COMB1 SLS Reactions

V2s V2s kN⋅:= V3s V3s kN⋅:= Ts Ts kN⋅ m⋅:=

NOMINAL EARTHQUAKE LOAD - EQX (R=1)

Frame Step Type

V2 SHEAR VERT

V3 SHEAR TRANS

T TORSION


TABLE: Element Forces - EQX

V2eqx V2eqx kN⋅:= V3eqx V3eqx kN⋅:= Teqx Teqx kN⋅ m⋅:=

NOMINAL EARTHQUAKE LOAD - EQY (R=1)

Frame Step Type

V2 SHEAR VERT

V3 SHEAR TRANS

T TORSION


TABLE: Element Forces - EQY

V2eqy V2eqy kN⋅:= V3eqy V3eqy kN⋅:= Teqy Teqy kN⋅ m⋅:=




Pier P6


MAXIMUM SHEAR FORCE DUE TO PLASTIC HINGING

The maximum shear force that the bearings will be subject to in the tranverse direction is theshear force due to plastic hinging at the base of the column.

The shear force due to plastic hingingat the top of the column tranverse direction VP 1197 kN⋅:=

This shear force should be carried by the transverse shear key on each bearing shelf or carried bythe bearings directly on on bearing shelf. The shear forces given above for earthquake loadingshould be used in the design if they are smaller than the plastic hinging effects.

MAXIMUM & MINIMUM VERTICAL ORRENPONDING PLASTIC HINGING EFECT

Pmax 5542kN:=

Pmin 1052− kN:=

Minimum Displacement Requirements (AASHTO LRFD Article 4.7.4.4)

Bridge seat widths at expansion bearings without restrainers, STU's or dampers shall eitheraccommodate the greater of the maximum displacement calculated from the seismic analysisor a percentage of the emprical seat width, N, specified below.

The percentage of N applicable to the bridge seismic zone shall be 150%.

The length of the bridge deckto the adjacent expansion joint Ld Ld m⋅:= Ld 73.6 m=

The height of the columns supporting the deck

H 9.912 m=

Skew of the supportmeasured from line normal to span

Sk 0=

The empirical seat width shall be taken as:

Ν 200 0.0017L

mm⋅+ 0.0067

Hmm⋅+⎛⎜

⎝⎞⎟⎠

1 0.000125 Sk2

⋅+⎛⎝

⎞⎠⋅ mm⋅:=

Ν 268 mm=

Check that the seat width provided is greater than 150% of N:

Seat width h5 1200 mm=

150% Ν⋅ 402 mm=

SeatWidth "OK" h5 Ν 150⋅ %≥if


:=

SeatWidth "OK"=




Pier P6


Design of Pier Coping based on plastic Hinge Force

Beam Column Joint

Bending moment due to plastic hinges force

Plastic Hinge Moment

MP 5542kN m⋅:=

Knee joints are the most common type of joint occuring in multicolumn bridge bents when tranverseresponse is considered. Equilibriun conditions under openinng and closing momentt are representedin figs. (a) and (b) above respectively. In these figures, the beam tensile, compressive, and shearstress resultants are indicated by Tb, Cb,and Vb, with Tc Cc, and Vcol being the corresponding force for the column. Axial forces Pc and Pbare present in the colun and beam, respectively. Moments Mb and Mc on joint boundaries induce theflexural stress resultants noted above.

Dimension :Column Beam Copinghc D:=hb v1:=

assume beam coping are rectangular for flexural design

b bf 2h5−:= de hb 100mm−:=




Pier P6



(a) Closing moment

Coping Dead Load at joint

Mdl MCop1:=

Mdl 20.6 kN m⋅=

MEU MP VPhb2

⋅+ Mdl+:=

MEU 7197.7 kN m⋅=

Strength reduction factor for flexure:

Φ 0.8=

Determine area of reinforcement, Af, in the coping to resist flexure(ref AASHTO LFRD Article 5.7.3.2.3 -see notes on flexural design for rearrangement of terms):

RMEU

Φ b⋅ de2

⋅ fy⋅:= R 0.0024= Μ

0.85 fc⋅

fy:= Μ 0.0654=

ρ Μ 1 12 R⋅Μ

−−⎛⎜⎝

⎞⎟⎠

⋅:= ρ 0.002424=

Af ρ de⋅ b⋅:=

Af 8930.5 mm2=


nbfAf

804 mm2⋅

:= nbf 11.1=

Provide 18 No 32mm φ bars in two layers np 18:=

Af np 804⋅ mm2⋅:=


β1 β1 0.85←

β1 β1 0.05fc 28MPa−


0.65 β1 0.65<if

:=

β1 0.836=


cAf fy⋅

0.85 fc⋅ β1⋅ b⋅:= c 189 mm=




Pier P6



cde

0.07=

MaxLimit "OK"cde

0.42≤if


:= MaxLimit "OK"=


fr 0.63fc



Scb de

2⋅

6:= Sc 1.616 m3

=




1.2 times the cracking moment Mcr•1.33 times the factored moment required by the applicable strength load combination•

MinimumSteel MrMEUΦ

←

M min 1.2Mcr 1.33 MEU⋅,( )←

"OK" Mr M≥if




Critical section for shear shall be taken as d v from the internal face of the support (AASHTOLRFD Article 5.8.3.2)

hs v1:=

dv 0.9 de⋅ 0.9 de⋅ 0.72 hs⋅>if


:=

dv 2368.8 mm=

Shear in coping due to plastic hinge force

VD2 MP⋅

dc

VPv12

⋅

dc+ Pmax+:= Pmax 5542 kN=

VD 7662 kN=




Pier P6



VEU VD VCop1+:=

VEU 8121.3 kN=

Section 1 range of 0 ~ 1 m from cl of column

VEU 8121 kN=


Vc 0.166fc


Vc 3015 kN=


Φs 0.7=


Vs VsVEUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:= Vs 8587 kN=




"FAIL" otherwise

:= CHECK "OK"=


φlink 25mm:=

Av πφlink

2

44⋅:= Av 1963 mm2

=


st st1Av fy⋅

0.0083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st 211 mm=




Pier P6



vuVEU

Φs b dv⋅( )⋅:= vu 3.498 MPa=




:=

smax 600 mm=



st otherwise

:=

st 211 mm=


VD2 MP⋅

dc

VPv12

⋅

dc+

56

Pmax+:=

VD 6738 kN=


VEU VD VCop2+:=

VEU 6735.9 kN=


Vc 0.166fc


Vc 3015 kN=


Φs 0.7=


Vs VsVEUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:= Vs 6607 kN=




Pier P6





"FAIL" otherwise

:= CHECK "OK"=


φlink 25mm:=

Av πφlink

2

44⋅:= Av 1963 mm2

=


st st1Av fy⋅

0.0083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st 275 mm=


vuVEU

Φs b dv⋅( )⋅:= vu 2.902 MPa=




:=

smax 600 mm=



st otherwise

:=

st 275 mm=


VD2 MP⋅

dc

VPv12

⋅

dc+

46

Pmax+:=

VD 5815 kN=




Pier P6



VEU VD VCop3+:=

VEU 5816.9 kN=


Vc 0.166fc


Vc 3015 kN=


Φs 0.7=


Vs VsVEUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:= Vs 5295 kN=




"FAIL" otherwise

:= CHECK "OK"=

Provide 19mmφ shear links wth 4 legs across the section: φlink 25mm:=

Av πφlink

2

44⋅:= Av 1963 mm2

=


st st1Av fy⋅

0.0083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st 343 mm=


vuVEU

Φs b dv⋅( )⋅:= vu 2.506 MPa=





Pier P6




:=

smax 600 mm=



st otherwise

:=

st 343 mm=

Design for Opening Moment and Closing Moment

MoC MP:= Plastic moment capacity

Vocol VP:= Plastic hinge shear force

hb v1:= Depth of beam

bje bf 2 h6⋅−( ):= Effective beam width

hc D:= Column width

Calculate Horizontal joint shear force and joint shear stress

Joint shear force

VjhMoChb

:=

Vjh 2028.6 kN=

Joint shear stress

vjVjh

bje hc⋅:=

vj 0.971 MPa=

CONVENTIONAL CAP BEAM DESIGN

Pc max Pmax Pmin,( ):= Max axial force corresponding plastic hinge effect

Pc 5542 kN=

Pb Vocol:=

The vertical and horizontal axial stress in the joint, allowing 45 degree spread into the cap beam , is

Vertical Horizontal

fvPc

bje hc 0.5 hb⋅+( )⋅:= fh

Pbbje hb⋅

:=

fv 1.183 MPa= fh 0.231 MPa=




Pier P6


Principal nominal stress

Compresion Tension

ptfv fh+

2

fv fh−

2

⎛⎜⎝

⎞⎟⎠

2

vj2

+−:=pcfv fh+

2

fv fh−

2

⎛⎜⎝

⎞⎟⎠

2

vj2

++:=

pt 0.374− MPa=pc 1.788 MPa=

pt 0.374 MPa=

The joint principal compression stress recommended limited to

plimitc 0.3fc:=

CHECK1 "OK" pc plimitc≤if

"DESIGN WELL-CONFINE JOINT" otherwise

:= CHECK1 "OK"=

The joint principal tension stress recommended limited to

plimitt 0.29fc

MPa⋅ MPa⋅:=

CHECK2 "OK" pt plimitt≤if

"DESIGN WELL-CONFINE JOINT" otherwise

:= CHECK2 "OK"=

Conclution notes : If principal tension stress is less than plimitt 1.588 MPa= , no vertical

joint reinforcement is needed, and only nominal tranverse hoop reinforcement is required.

Design for Joint Reinforcement

Closing momentdc 60mm:= Clear cover to longitudinal bar

nc 24:= Number of column longitudinal reinforcement

φbar 32mm:= Bar diameter

fy 390 MPa= Joint reinforcement nominal yield

fye 1.1fy:= Steel design yield strength

foyc 1.4 fy⋅:= Steel yield overstress

Asc nc φbar2

⋅π

4⋅:= Area of column reinforcement

la 2350mm:= Main column reinforcement ajjacent to beam

Dr D 2 dcφbar

2+

⎛⎜⎝

⎞⎟⎠

⋅−:=

fsh 0.0015 Es⋅:=




Pier P6


Volumetric ratio of tranverse hoop reinforcement for Low principal tension stress

Minimum tranverse hoop reinforcement

ρsmin

0.29fc

MPa⋅ MPa⋅

fy:=

ρs10.46 Asc⋅ foyc⋅

Dr la⋅ fsh⋅:=

ρsmin 0.00407=ρs1 0.0072536=

ρs max ρs1 ρsmin,( ):=

Calculate hoops reinforcement

φh 19mm:= hoop reinforcement diameter

Ahπ φh

2⋅

4:=

s4 Ah⋅

ρs Dr⋅:= s 164.928 mm=

Opening joint

Anchorage to column bars closest to the cap beam is provided by struts D1, directed toward thecolumn compression resultant Cc, nd D2 directed outward the column into the beam. The verticalcomponent D2, namely Ts, is provided by stirrups close to joint. Tranfers of this tension force to the top of the beam provides the neccesary force to incline the beam compression force Cb into majorcompression arch D3. Horizontal component of D1 and D2 approximately balance each other, reducing the need for hoop reinforcement. It is recommended that with this design, 50% of Tc (that portion closest to the inner face of the column)be tranferred by this mechanism. Allowing 50% of this force to be tranferred into to the joint via D2, the tension force Ts in external beam stirrups is Ts = 0.25Tc. Again, approximating Tc = 0.5AscFoyc, the required amount of vertical beam stirrup reinforcement is

fyv fy:=

Ajv 0.25 Asc⋅foycfyv

⋅:=

Ajv 6755.68 mm2=




Pier P6


Amount vertical reinforcement Ajv 6755.68 mm2= to be placed over a length not greater than

0.5 hb⋅ 1366 mm= this is can provide :

φv 25mm:=

Avπ φv

2⋅

4:=

nvAjvAv

:=

nv 13.763=

This can be provided by 14 D25 in 7 layer of 2 stirrup leg each.

The strut D2 imposes additional tension force in the beam bottom flrxural reinforcement, as isapparent from equillibrium of force under D2 an Ts. Assuming the special vertical reinforcement to beplace over length 0.5 hb from face of column as calculated above, the additional horizontal force to beresisted by the bottom beam reinforcement will be approximatelly ).5Ts, the additionall area of beambottom reinforcement required is thus

fyb fy:=ΔAsb 0.0625Asc

foycfyb

⋅:= where :

ΔAsb 1688.9 mm2=

nsbΔAsbAv

:=

nsb 3.441=

To provide assistance in bond tranfer on top reinforcement and to avoid the total beam tension forcebeing tranferred across to the hook, it is recommended that vertical stirrups, inside the column core,be provided for a vertical resisttance equal to 0.5 Ts. this requires an internal vertical joint stirrup area of:

Avi 0.0625Ascfoycfyb

⋅:=

Avi 1689 mm2=

nviAviAv

:=

nvi 3.441=




Pier P6


Design of Cantilever Slab

Equivalent Width of Deck Overhang (AASHTO LRFD Table 4.6.2.1.3-1)

Width of coping slab at support:

b bf 2 h5⋅−:= b 1400 mm=

Distance from outermost load to point of support:

X h4 300mm− 350mm−( ):= X 1510 mm=

Equivalent width of deck overhang:

bew bew 570mm 0.416 X⋅+←

bew bew b≤if

b otherwise

:= bew 1198 mm=

Check Deflection of Cantilever Slab (AASHTO LRFD Article 2.5.2.6.2)

Depth of cantilever slab in main deck section at support:

hds 450mm:=

Span length of cantilever slab in main deck section:

lds 2645mm:=

Depth of deck overhang at support to match main deck outline:

hcs1 hds 250mm−( ) h4lds⋅ 250mm+:= hcs1 413 mm=

Depth of deck overhang at outermost load point:

hcs2 hds 250mm−( ) 350mm 300mm+

lds⋅ 250mm+:= hcs2 299 mm=




Pier P6


Depth of deck overhang at innermost load point:

hcs3 hcs3 hds 250mm−( ) 350mm 300mm+ 1750mm+

lds⋅ 250mm+←

hcs3 hcs3 hcs1≤if

hcs1 otherwise

:=

hcs3 413 mm=

Distance from innermost load to point of support:

X3 X3 h4 300mm− 350mm− 1750mm−←

X3 X3 0>if

0m otherwise

:= X3 0 mm=

Moment of inertia of equivalent width of deck overhang assuming cracked section:

at support Iew1 40 %⋅bew hcs1

3⋅

12⋅:=

at outer load point Iew2 40 %⋅bew hcs2

3⋅

12⋅:=

at innermost load point Iew3 40 %⋅bew hcs3

3⋅

12⋅:=

The deflection under the applied wheel loads is then given by (increaseed by 30% to account fordynamic loading):

δT112.5 kN⋅ 130⋅ %

Ec

0

X

xx2

Iew2 Iew1 Iew2−( ) xX⋅+

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦

⌠⎮⎮⎮⎮⌡

d

0

X3

xx2

Iew3 Iew1 Iew3−( ) xX3⋅+

⎡⎢⎢⎢⎣

⎤⎥⎥⎥⎦

⌠⎮⎮⎮⎮⌡

d+

⎡⎢⎢⎢⎢⎣

⎤⎥⎥⎥⎥⎦

⋅:=

δT 2.61 mm=

Check that the deflection does not exceed the limit for vehicular load on cantilever arms:

δLIMITX

300:= δLIMIT 5.03 mm=

DEFLECTIONCHECK "OK" δT δLIMIT≤if

"FAIL" otherwise

:=

DEFLECTIONCHECK "OK"=




Pier P6



Nominal bending moment in slab at support

Coping self weight - cantilever wings:

wcw0.25m 0.45m+

2bf h5 2⋅−( )⋅ 24.5

kN

m3⋅:= wcw 12.0

kNm

=

Mcw wcwh42

2⋅:= Mcw 28.0 kN m⋅=

Railing dead load - each side:

Wr0.433

2m2 bf h5 2⋅−( )⋅ 24.5

kN

m3⋅:= Wr 7.4 kN=

Railing weight Mr Wr h4 0.15m+ 0.25m−( )⋅:= Mr 15.3 kN m⋅=

Superimposed deadload on coping

wsdl 0.125m bf h5 2⋅−( )⋅ 22.5kN

m3⋅:= wsdl 3.9

kNm

=

Msdl wsdlh4 350mm−( )2

2⋅:= Msdl 6.4 kN m⋅=


Mt 112.5 kN⋅ X X3+( )⋅:= Mt 169.9 kN m⋅=

Total bending moment in slab at face of support - Ultimate Limit State

MSU Mcw Mr+( ) 1.3⋅ Msdl 2.0⋅+ Mt 1.3⋅( ) 1.8⋅+:=

MSU 466.7 kN m⋅=

Depth of section:

hcs hcs1:= hcs 413 mm=

Effective depth of slab:

de hcs 90mm−( ):= de 323 mm=

Effective width of coping slab at support:

bew 1198 mm=

Strength reduction factor for flexure: Φ 0.8=




Pier P6


Determine area of reinforcement, Af, in the coping to resist flexure(ref AASHTO LFRD Article 5.7.3.2.3 - see notes on flexural design for rearrangement of terms):

RMSU

Φ bew⋅ de2

⋅ fy⋅:= R 0.0119= Μ

0.85 fc⋅

fy:= Μ 0.0654=

ρ Μ 1 12 R⋅Μ

−−⎛⎜⎝

⎞⎟⎠

⋅:= ρ 0.013294=

Af ρ de⋅ b⋅:= b 1400 mm=

Af 6017.5 mm2=


nbfAf

804 mm2⋅

:= nbf 7.5=


nbars 10:= Af nbars 804⋅ mm2⋅:= Af 8040 mm2

=


β1 β1 0.85←

β1 β1 0.05fc 28MPa−


0.65 β1 0.65<if

:=

β1 0.836=


cAf fy⋅

0.85 fc⋅ β1⋅ b⋅:= c 105 mm=


cde

0.33=

MaxLimit "OK"cde

0.42≤if


:= MaxLimit "OK"=


fr 0.63fc



Scb hcs

2⋅

6:= Sc 0.04 m3

=






Pier P6




MinimumSteel MrMSUΦ

←

M min 1.2Mcr 1.33 MEU⋅,( )←

"OK" Mr M≥if




The depth of concrete in compression, C, at the serviceability limit state, assuming one levelof rebar, is given by:

CAf

2 2 b⋅ Af⋅ de⋅

α+ Af−

bα⋅:= C 128 mm=

Calculate lever arm, z:

z deC3

−:= z 281 mm=

Total bending moment in slab at face of support - Service Limit State

MSS Mcw Mr+( ) 1.0⋅ Msdl 1.0⋅+ Mt 1.3⋅( ) 1.0⋅+:=

MSS 270.6 kN m⋅=

Calculate the maximum stress in the reinforcement at SLS:

fsMSSAf z⋅

:= fs 120 MPa=



mm⋅:=


Area of concrete with same centoidper bar A

b dc⋅ 2⋅

nbars:=

fsa fsaZ

dc A⋅( )1

3

←

fsa fsa 170MPa<if


:=

fsa 170 MPa=







Pier P6



Total force in rebar Stress in concrete

TS Af( ) fs⋅:= TS 964 kN= fsc fsC

de C−⋅

1α⋅:= fsc 10.8 MPa=

Force in concrete

CC fscb C⋅2

⋅:= CC 964 kN=



dv 0.9 de⋅ 0.9 de⋅ 0.72 hcs⋅>if

0.72 hcs⋅ otherwise

:=

dv 298 mm=

Shear in coping slab at critical section due to loads applied on coping body


Vcw wcw h4 dv−( )⋅:= Vcw 22.4 kN=

Railingweight

Vr Wr:= Vr 7.4 kN=


Vsdl wsdl h4 600mm− dv−( )⋅:= Vsdl 5.0 kN=


Vt 112.5 kN⋅:= Vt 112.5 kN=

Total shear force in coping slab at face of support - Ultimate Limit State


VSU Vcw Vr+( ) 1.3⋅ Vsdl 2.0⋅+ Vt 1.4⋅( ) 1.8⋅+:=

VSU 332.2 kN=


Vc 0.166fc


Vc 379 kN=


Φs 0.7=




Pier P6



Vs VsVSUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:= Vs 96 kN=




"FAIL" otherwise

:= CHECK "OK"=

Provide 13mmφ shear links with 4 legs across the section

φlink 13mm:=

Av πφlink

2

44⋅:= Av 531 mm2

=


st st1Av fy⋅

0.0083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st 644 mm=


vuVSU

Φs b dv⋅( )⋅:= vu 1.139 MPa=




:=

smax 238 mm=



st otherwise

:= st 238 mm=

Provide 13mmφ links at 150c/c




Pier P6


Beam Ledge/Corbel Design AASHTO LRFD Article 5.13.2.5

General, Article 5.13.2.5.1

As illustrated below, beam ledges shall resist:

Flexure, shear and horizontal forces at the loaction of Crack 1;•Tension force in the supporting element at the location of Crack 2;•Punching shear at points of loading at the location of Crack 3; and•Bearing forces at the location of Crack 4•




Pier P6


Design for Shear, Article 5.13.2.5.2

Design of beam ledges for shear shall be in accordance with the requirements of shear friction inArticle 5.8.4.

The width of the concrete face assumed to participate shall not exceed the width ilustrated below:

Edge distance of bearing C h2:= C 800 mm=



Width of the interface bv 2 C⋅:= bv 1600 mm=



Loads on Ledge

PC Deck Dead Load Reaction - max at the bearing - ULS Comb1

V2upc

V2u2

2

Tu2

h1c 2⋅+:= V2upc 3512.7 kN=

Steel Deck Dead Load Reaction - max at the bearing - ULS Comb1

V2ust

V2u3

2

Tu3

h1s 2⋅+:= V2ust 3232.7 kN=




Pier P6


Ultimate shear force in coping from max bearing reactions at face of column:

PC deck reaction Vupc V2upc:= Vupc 3512.7 kN=

Steel deck reaction Vust V2ust:= Vust 3232.7 kN=

Maximum design shear force

Vu max Vupc Vust,( ):= Vu 3513 kN=



mm2⋅ N⋅<if

5.5Acv


:=

Vnlimit 9988 kN=

Calculate required nominal shear strength and check beam ledge depth

VnVuΦs

:= Vn 5018.2 kN=

BeamLedge "OK" Vn Vnlimit≤if


:=BeamLedge "OK"=



c 1.0MPa:= λ 1.00:= μ 1.4 λ⋅:=



fy μ⋅←

Avf2 0.35bvmm⋅

MPafy

⋅ mm2⋅←



0VnAcv

0.7MPa<if

:=

VnAcv

2.763 MPa=

Avf 5864.8 mm2=




Pier P6


Design for Flexure and Horizontal Force, Article 5.13.2.5.3

The area of total primary tension reinforcement shall satisfy the requirements of Article 5.13.2.4.2.

The primary tension reinforcement shall be spaced uniformly with the region 2C.

The section at the face of the support shall be designed to resist simultaneously a factored shearforce Vu, a factored moment Mu and a concurrent factored horizontal tensile force Nuc.

Nuc shall not be taken to be less than 0.2Vu and shall be regarded as a live load, even where itresults from creep, shrinkage or temperature change.

These provisions apply to beam ledges:

• with a shear span-to-depth ratio av/de not greater than unity

• subject to a horizontal tensile force Nuc not larger than Vu

The depth at outside edge of bearing shall not be less than 0.5de, where d e is effective depth.

Horizontal tensile force Nuc 0.2 Vu⋅:= Nuc 702.5 kN=

Shear span avh52

:= av 600 mm=

Design Moment Mu Vu av⋅ Nuc h de−( )⋅+:= Mu 2153.3 kN m⋅=



Width of section b 2 C⋅:= b 1600 mm=


RMu

Φ b⋅ de2

⋅ fy⋅:= R 0.0033= Μ

0.85 fc⋅

fy:= Μ 0.0654=

ρ Μ 1 12 R⋅Μ

−−⎛⎜⎝

⎞⎟⎠

⋅:= ρ 0.003439=

As ρ b⋅ de⋅:= As 6244.9 mm2=


AnNucΦ fy⋅

:= An 2252 mm2=




Pier P6



As As As23

Avf⋅ An+>if

23


:=



As otherwise

:=

As 6245 mm2=

AsAsb

:= As 3903mm2

m= As 3.903

mm2

mm=


ρREQUIREDAs m⋅



Ah 0.5 As b⋅ An−( )⋅:= Ah 1997 mm2=


Anchorage of primary reinforcement:

At the front face of the beam ledge, the primary tension reinforcement, As, shall be anchoredby one of the following:

a) a structural weld to a transverse bar of at least equal size; weld to be designed to developspecified yield strength fy of As bars

b) bending primary tension bars As back to form a horizontal loop

c) some other means of positive anchorage

The bearing area of load on the bracket or corbel shall not project beyond interior face of transverseanchor bar (if one is provided).




Pier P6


Design for Punching Shear, Article 5.13.2.5.4

The truncated pyramids assumed as failure surfaces for punching shear, as illustrated below,shall not overlap.

Applied ultimate reaction Vu 3513 kN=

Width of bearing L 800 mm=

Length of bearing W 800 mm=

Effective depth de 1135 mm=

Bearing pad spacing S min h1c h1s,( ) 2⋅:= S 6150 mm=

Nominal punching shear resistance, Vn, shall be taken as:At interior pads, or exterior pads, where the end distance C is greater than half the pad spacing•S/2:

Vn1 0.328fc

MPa⋅ MPa⋅ W 2 L⋅+ 2 de⋅+( )⋅ de⋅:= Vn1 9522 kN=

At exterior pads where the end distance C is less than half the pad spacing S/2 and C-0.5W•is less than de:

Vn2 0.328fc

MPa⋅ MPa⋅ W L+ de+( )⋅ de⋅:= Vn2 5577 kN=

At exterior pads where the end distance C is less than half the pad spacing S/2 but C-0.5W•is greater than de:

Vn3 0.328fc

MPa⋅ MPa⋅ 0.5W L+ de+ C+( )⋅ de⋅:= Vn3 6392 kN=




Pier P6


Determine the nominal punching resistance:

Vn Vn1 CS2

≥if

Vn2 CS2

<⎛⎜⎝

⎞⎟⎠

C 0.5 W⋅−( ) de≤⎡⎣ ⎤⎦⋅if

Vn3 otherwise

:=

Vn 5577 kN=

Check that the nominal punching resistance is adequate:

PunchingShearCHECK "SATISFIED" Vn Φs⋅ Vu≥if

"FAIL" otherwise

:=

PunchingShearCHECK "SATISFIED"=

Check Torsional Requirements (AASHTO LRFD Section 5.8)

Check the torsional moment requirements of the beam ledge assuming the bearings (on the steeldeck side) are fully loaded and the opposite bearings (on the PC deck side) are loaded only withpermanent load:


V2pc

V2dl22

Tdl2h1c 2⋅

+:=

V2pc 752.9 kN=

Superimposed Dead Load PC Deck Reaction - max at the bearing

V2st

V2dl32

Tdl3h1s 2⋅

+:=

V2sdl2290 kN=

Ultimate limit state reaction in bearing under permanent load:

Vupc V2pc 1.3⋅ V2sdl22.0⋅+:=

Vupc 1558 kN=

Calculate torsion in the coping:

Tc Vu Vupc−( )bf2

h52

−⎛⎜⎝

⎞⎟⎠

:=

Tc 2541 kN m⋅=

Total shear force in coping at face of column - Ultimate Limit State


VCU Vc Vcw+ Vr+( ) 1.3⋅ Vsdl 2.0⋅+:=

VCU 541.1 kN=




Pier P6


Associated shear force:

VTU VCU Vust+ Vupc+:=

VTU 5332 kN=

The coping will resist torsion moment with two torson blocks as illustrated below. The torsionmoment will be distributed into each torsion block in accordance with the relative area of eachblock.

Dimensions of the torsion blocks are as follows:

Torsion block 1 h1 v1:= h1 2732 mm=

b1 bf h5 2⋅−:= b1 1400 mm=

Torsion block 1 h223

v3⋅:= h2 800 mm=

b2 bf:= b2 3800 mm=

Area enclosed with centerline of transverse rebar



Area enclosed by shear flow path



Calculate torsion moments carried by each block:


Ao1 Ao2+Tc⋅:= Tc1 1443 kN m⋅=


Ao1 Ao2+Tc⋅:= Tc2 1098 kN m⋅=




Pier P6



Depth of section:

hs v1:=


de hs 150mm−( ):= de 2582 mm=

Width of coping:

b bf h5 2⋅−:= b 1400 mm=


dv 0.9 de⋅ 0.9 de⋅ 0.72 hs⋅>if


:=

dv 2323.8 mm=


Vc 0.166fc


Vc 2958 kN=


Φs 0.7=


Vs VsVTUΦs

Vc−←

Vs Vs 0kN>if

0kN otherwise

:=

Vs 4659 kN=


φlink 19mm:=

Av πφlink

2

44⋅:= Av 1134 mm2

=


st1 st1Av fy⋅

0.0083fc

MPa⋅ MPa⋅ b⋅

←

st2Av fy⋅ dv⋅

Vs← Vs 0kN>if


st1 otherwise

:= st1 221 mm=




Pier P6


For reinforced concrete the angle of inclination of diagonal stresses, θ, can be assumed tobe 45 degrees:

θ 45deg:=

Determine required transverse reinforcement for torsion:


φlink2

4:= At 284 mm2

= φlink 19 mm=

Required spacing of torsional reinforcement st2

2 Ao1⋅ At⋅ fy⋅ cot θ( )⋅


Calculate combined spacing of shear and torsion transverse reinforcement:

1st1

1st2

+⎛⎜⎝

⎞⎟⎠

1−129.303 mm=

Provide 19mm dia transverse reinforcement at 100mm c/c in main body of coping



φlink 16mm:=

Av πφlink

2

42⋅:= Av 402 mm2

=

Determine required transverse reinforcement for torsion:


φlink2

4:= At 201 mm2

= φlink 16 mm=

Required spacing of torsionalreinforcement

st22 Ao2⋅ At⋅ fy⋅ cot θ( )⋅


Provide 16mm dia transverse reinforcement at 200mm c/c in beam ledges




Pier P6


Design for Loads from Deck Jacking

AASHTO LRFD requires that the beam ledge is designed to resist deck jacking forces.

The deck jacking loads shall not be less than 1.3 times the permanent load reaction at the bearingadjacent to the point of jacking.


Vdlpc

V2dl22

Tdl2h1c 2⋅

+:= Vdlpc 752.9 kN=


Vdlst

V2dl32

Tdl3h1s 2⋅

+:= Vdlst 665.9 kN=

PC Deck Superimposed Dead Load Reaction - max at the bearing

Vsdlpc

V2sdl22

Tsdl2h1c 2⋅

+:= Vsdlpc 144.9 kN=

Steel Deck Superimposed Dead Load Reaction - max at the bearing

Vsdlst

V2sdl32

Tsdl3h1s 2⋅

+:= Vsdlst 121.2 kN=

Maximum design load reaction due to deck jacking:

VJack 1.3 max Vdlpc Vsdlpc+ Vdlst Vsdlst+,( ):=

VJack 1167 kN=

Assuming the worst case for positioning of the jack and ignoring width of loaded area from jack gives:

Horizontal tensile force Nuc 0.2 VJack⋅:= Nuc 233.4 kN=

Shear span av h5:= av 1200 mm=

Design Moment Mu VJack av⋅ Nuc h de−( )⋅+:= Mu 1078.0 kN m⋅=

Loaded width during jacking w 1000mm:= w 1000 mm=



Width of the interface bv w:= bv 1000 mm=






Pier P6


Calculate required nominal shear strength

VnVJackΦs

:= Vn 1667.4 kN=



c 1.0MPa:=

λ 1.00:=

μ 1.4 λ⋅:=



fy μ⋅←

Avf2 0.35bvmm⋅

MPafy

⋅ mm2⋅←



0VnAcv

0.7MPa<if

:=

VnAcv

1.469 MPa=

Avf 975 mm2=



Width of section b 1000mm:=


RMu

Φ b⋅ de2

⋅ fy⋅:= R 0.0027= Μ

0.85 fc⋅

fy:= Μ 0.0654=

ρ Μ 1 12 R⋅Μ

−−⎛⎜⎝

⎞⎟⎠

⋅:= ρ 0.002739=

As ρ b⋅ de⋅:= As 3109.2 mm2=


AnNucΦ fy⋅

:= An 748 mm2=




Pier P6



As As As23

Avf⋅ An+>if

23


:=



As otherwise

:=

As 3492 mm2=

AsAsb

:= As 3492mm2

m=


ρREQUIREDAs m⋅



Ah 0.5 As b⋅ An−( )⋅:= Ah 1372 mm2=


CONCLUSION - PROVIDE SAME REINFORCEMENT FOR MAIN BEAM LEDGE DESIGN ACROSSENTIRE BEAM LEDGE WIDTH - EXCEPT OVER WIDTH OF COLUMN.

JACKING POINTS SHALL NOT BE CLOSER THAN 500mm FROM FACE OF COLUMN.




Pier P6


Design for Loads from Longitudinal Restrainers, Article 3.10.9.5

This calculation note is intended for the design of the reinforced concrete elements of the piercoping supporting the restrainers.

For the design of the restrainers themselves, inlcuding any local bursting reinforcement required,refer to a separatee calculation.

Restrainers shall be designed for a force calculated as the acceleration coefficient times thepermanent load of the lighter of the two adjoining spans.Acceleration coefficient

A 0.40:=

Permanent load of the lighter of the two spans assuming, conservatively that only 40% of thetotal load appears as a reaction at the pier coping:

PERMLOAD V2pc V2sdl2+⎛

⎝⎞⎠

140%

:= PERMLOAD 2606 kN=

Design load in restrainer

RESTLOAD PERMLOAD A⋅:= RESTLOAD 1042 kN=

Required nominal shear strength of pier coping supporting the restrainers:

VnrRESTLOAD

Φs:= Vnr 1489 kN=

Assume conservatively that this load is carried at a single point with a 150mm x 150mm bearingplate located at mid height of the PC deck. Given that the applied load will be supported by acoping width that is at least equal to the shear span, av, of the load - design shear frictionreinforcement in accordance with Article 5.13.2.5.2 and Article 5.8.4.




Pier P6


Shear span of the load av v1 v2−1200mm

2−:= av 932 mm=

The width of the concrete face assumed to participate in resistance to shear is as defined below:

bv S min 2 h1c⋅ 2 h1s⋅,( )←

dw 150mm 4 av⋅+←

dw dw S<if

S otherwise

:= bv 3878 mm=

Depth of support h bf h5 2⋅−:= h 1400 mm=






mm2⋅ N⋅<if

5.5Acv


:= Vnlimit 28474.2 kN=

Check that the required nominal shear strength is less than that provided

PierCopingCapacity "OK" Vnr Vnlimit≤if


:=PierCopingCapacity "OK"=



c 1.0MPa:= λ 1.00:= μ 1.4 λ⋅:=


Avf Avf1Vnr c Acv⋅−

fy μ⋅←

Avf2 0.35bvmm⋅

MPafy

⋅ mm2⋅←



0VnrAcv

0.7MPa<if

:=

VnrAcv

0.288 MPa=

Avf 0 mm2=

Conclude that the concrete section alone has adequate capacity to carry the loads from thelongitudinal restrainers without the need for addiional shear reinforcement.


detailed design toc - jicaultimate moment capacity of reinforced concrete beams is determined in...

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